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All Activities in Section 5, Work and Energy
Section 8-1:
Conservative and
Nonconservative Forces
Problem
1. Determine the work done by the frictional force in moving a block of mass m from point 1 to point 2 over the two paths shown in Fig. 8-26. The coefficient of friction has the constant value µ over the surface. (The diagram lies in a horizontal plane.)
2
(b)
1
(a)
figure8-26 Problems 1, 2.Solution
Figure 8-26 is a plane view of the horizontal surface over which the block is moved, showing the paths (a) and (b). The force of friction is µmg opposite to the displacement (f · dr = −fdr), so W(a) =
−µmg(ℓ + ℓ) = −2µmgℓ, and W(b)= −µmg
√
ℓ2+ ℓ2=
−√2µmgℓ. Since the work done depends on the path, friction is not a conservative force.
Problem
2. Now take Fig. 8-26 to lie in a vertical plane, and find the work done by the gravitational force as an object moves from point 1 to point 2 over each of the paths shown.
Solution
Take the origin at point 1 in Fig. 8-26 with the x-axis horizontal to the right and the y-axis vertical upward. The gravitational force on an object is constant, Fg=
−mgˆ, while the paths are (a) dr =ˆdy for x = 0 and 0 ≤ y ≤ ℓ, followed by dr =ˆıdx for y = ℓ and
0 ≤ x ≤ ℓ, and (b) dr =ˆıdx +ˆdy = (ˆı+ˆ)dy, for 0 ≤ y ≤ ℓ (since x = y along this path). The work done by gravity (Equation 7-11) is
Wg(a)= Z Fg· dr = Z ℓ 0 (−mgˆ) ·ˆdy + Z ℓ 0 (−mgˆ) ·ˆıdx = −mg Z ℓ 0 dy + 0 = −mgℓ, and Wg(b)= Z ℓ 0 (−mgˆ) · (ˆı+ˆ)dy = −mg Z ℓ 0 dy = −mgℓ.
Of course, these must be the same because gravity is a conservative force.
Problem
3. The force in Fig. 8-22a is given by F = F0ˆ, where
F0 is a constant. The force in Fig. 8-22b is given by
F =F0(x/a)ˆ, where the origin is taken at the lower
left corner of the box, a is the width of the square box, and the distance x increases horizontally to the right. Determine the work done by F on an object moved counterclockwise around each box, starting at the lower left corner.
(a) (b)
figure8-22 Problem 3.
Solution
The path around the square consists of four segments, each along a side, in the direction of a
counterclock-wise circulation. Thus W = I F · dr = Z a 0 F(y = 0) ·ˆıdx + Z a 0 F(x = a) ·ˆdy + Z 0 a F(y = a) ·ˆıdx + Z 0 a F(x = 0) ·ˆdy (Note that a path parallel to the x-axis from right to left is represented by dr =ˆıdx with x going from a to 0 in the limits of integration, etc.) (a) For F = F0ˆ,
the expression for the work becomes W = 0 + F0 Z a 0 dy + 0 + F0 Z 0 a dy = F0a − F0a = 0.
(b) For F = F0(x/a)ˆ, the work around the square is
W = 0 + F0 a a Z a 0 dy + 0 + 0 = F0a.
Section 8-2:
Potential Energy
Problem
4. Rework Example 8-1, now taking the zero of potential energy at street level.
Solution
(a) The office in question is 32 stories above the street level (the first floor) where U1= 0, so the difference
in gravitational potential energy is ∆U = U33− U1=
U33− 0 = mg∆y = (55×9.8 N)(32×3.5 m) = 60.4 kJ.
(b) At the fifty-ninth floor, U59− U1= (55×9.8 N)×
(58×3.5 m) = 109 kJ. (c) Street level is the zero of potential energy, U1= 0. Note that the differences in
potential energy between any two levels are the same as in Example 8-1, e.g., U59− U33= (109 − 60.4) kJ =
49.0 kJ.
Problem
5. Find the potential energy of a 70-kg hiker (a) atop New Hampshire’s Mount Washington, 1900 m above sea level, and (b) in Death Valley, California, 86 m below sea level. Take the zero of potential energy at sea level.
Solution
If we define the zero of potential energy to be at zero altitude (y = 0), then U (0) = 0, and Equation 8-3 (for the gravitational potential energy near the surface of the Earth, |y| ≪ 6370 km) gives U(y) − U(0) = U (y) = mg(y − 0) = mgy. Therefore, (a) U(1900 m) = (70×9.8 N)(1900 m) = 1.30 MJ, and (b) U(−86 m) = (70×9.8 N)(−86 m) = −59.0 kJ.
Problem
6. An incline makes an angle θ with the horizontal. Find the gravitational potential energy associated with a mass m located a distance x measured along the incline. Take the zero of potential energy at the bottom of the incline.
Solution
To avoid confusion, let x-y refer to the Earth’s surface, and x′-y′ refer to the incline. The gravitational
potential energy (g assumed constant, zero at y = 0) is U = mgy. For a point on the incline, y = x′sin θ, so
U = mgx′sin θ.
Problem 6 Solution.
Problem
7. Show using Equation 8-2b that the potential energy difference between the ground and a distance h above the ground is mgh regardless of whether you choose the y-axis upward or downward.
Solution
Equation 8-2b gives the potential energy difference for a constant force in the y direction, ∆U =
U (y2) − U(y1) = −Fy(y2− y1). If you take the y-axis
upward, with the ground at y1, the gravitational force
is Fy = −mg, while a point a distance h above ground
is y2= h + y1. Then ∆U = −(−mg)h = mgh. On the
other hand, if the y-axis is downward, then Fy= mg
but y2= −h + y1, so ∆U = −(mg)(−h) = mgh is the
same.
Problem
8. The top of the volcano Haleakala on Maui, Hawaii, is 3050 m above sea level and 18 km inland from the sea. By how much does your gravitational potential energy change as you come down from the mountain-top observatory to swim in the ocean? Assume your mass is 75 kg.
Solution
U (sea level) − U(mt.top) = −mgh = −(75 kg)× (9.8 m/s2)(3050 m) = −2.24 MJ.
Problem
9. A 1.50-kg brick measures 20.0 cm×8.00 cm× 5.50 cm. Taking the zero of potential energy when the brick lies on its broadest face, what is the potential energy (a) when the brick is standing on end and (b) when it is balanced on its 8-cm edge, with its center directly above that edge? Note: You can treat the brick as though all its mass is
concentrated at its center.
Solution
The center of the brick is a distance ∆y = 10 cm − 2.75 cm = 7.25 cm above the zero of potential energy in position (a), and ∆y = 1
2p(20 cm)2+ (5.5 cm)2 −
2.75 cm = 10.4 cm − 2.75 cm = 7.62 cm in position (b) (see sketch; the center is midway along the diagonal of the face of the brick). From Equation 8-3, the gravitational potential energy is Ua= mg ∆y =
(1.5×9.8 N)(7.25 cm) = 1.07 J and Ub = (1.5×
9.8 N)(7.62 cm) = 1.12 J above the zero energy.
Problem 9 Solution.
Problem
10. A 60-kg hiker ascending 1250-m-high Camel’s Hump mountain in Vermont has potential energy −2.4×105 J; the zero of potential energy is taken
at the mountain top. What is her altitude?
Solution
If we measure the altitude from sea level, U (y) − U (1250 m) = mg(y − 1250 m) = −2.4×105J, or
y = 1250 m − 2.4×105 J/(60 kg)(9.8 m/s2) = 842 m.
Problem
11. How much energy can be stored in a spring with k = 320 N/m if the maximum allowed stretch is 18 cm?
Solution
From Equation 8-4, U = 1 2kx 2=1 2(320 N/m)×(0.18 m)2= 5.18 J is the maximum potential energy of
Problem 10 Solution.
the spring whose stretch is not greater than 18 cm.
Problem
12. A carbon monoxide molecule can be modeled as a carbon atom and an oxygen atom connected by a spring. If a displacement of the carbon by
1.6×10−12 m from its equilibrium position relative
to the oxygen increases the molecule’s potential energy by 0.015 eV, what is the spring constant?
Solution
In this model, the zero of potential energy is at the equilibrium separation for the molecule, so the spring constant can be calculated from Equation 8-4 and the energy, k = 2U/x2= 2(0.015 eV)(1.6×10−19 J/eV)÷
(1.6×10−12 m)2= 1.88 kN/m.
Problem
13. How far would you have to stretch a spring of spring constant k = 1.4 kN/m until it stored 210 J of energy?
Solution
Assuming one starts stretching from the unstretched position (x = 0), Equation 8-4 gives x =p2U/k = p2(210 J)/(1.4 kN/m) = 54.8 cm.
Problem
14. A more accurate expression for the force law of the rope in Example 8-3 is F = −kx + bx2− cx3,
where k and b have the values given in
Example 8-3, and c = 3.1 N/m3. Find the energy stored in stretching the rope 2.62 m. By what percentage does your result differ from that of Example 8-3?
Solution
The more accurate expression for force adds a term − ∫02.62 cm(−cx3)dx = 14cx4 2.62 m 0 = 1 2(3.1 N/m 3 )×
(2.62 m)4= 36.5 J to the stored potential energy,
which is about 4.9% of the 741 J of potential energy found in Example 8-3.
Problem
15. The force exerted by an unusual spring when it’s compressed a distance x from equilibrium is given by F = −kx − cx3, where k = 220 N/m, and c =
3.6 kN/m3. Find the energy stored in this spring when it’s been compressed 15 cm.
Solution
For a one-dimensional force, one can use Equation 8-2a to find U (x) − U(0) = − ∫015 cm(−kx − cx3)dx = 12kx2+14cx4 15 cm 0 = 1 2(220 N/m)(0.15 m) 2+ 1 4(3.6 kN/m 3
)(0.15 m)4= 2.93 J. Since this energy
could be recaptured from the spring, in the form of kinetic energy imparted to a mass, for example, it is referred to as stored energy.
Problem
16. The force on a particle is given by F = Aˆı/x2,
where A is a positive constant. (a) Find the potential energy difference between two points x1
and x2, where x1> x2. (b) Show that the
potential energy difference remains finite even when x1→ ∞.
Solution
(a) U (x2) − U(x1) = − Z x2 x1 A x2dx = A 1 x x2 x1 = A 1 x2 − 1 x1 .(b) For x1→ ∞, U(x2) − U(∞) = A/x2. In this case,
it makes sense to define the zero of potential energy at infinity, U (∞) = 0, so U(x) = A/x.
Problem
17. A particle moves along the x-axis under the influence of a force F = ax2+ b, where a and b are
constants. Find its potential energy as a function of position, taking U = 0 at x = 0.
Solution
Equation 8-2a, with U (0) = 0, gives U (x) = − Z x 0 Fxdx′= − Z x 0 (ax′2+ b)dx′ = −1 3ax 3− bx.
Problem
18. A 3.0-kg fish is hanging from a spring scale whose spring constant is 240 N/m. (a) What is the potential energy of the spring? (b) If the fish were moved slowly upward to the equilibrium position of the spring, by how much would its gravitational potential energy change? (c) In case (b), by how much would the spring’s potential energy change? Explain any apparent discrepancies.
Solution
To avoid confusion, call the place where the fish hangs at rest the equilibrium position (B), and the place where the spring has its natural length the unstretched position (A). (a) At equilibrium (B), kx = mg, so the potential energy of just the spring is Us(B) = 12kx2= 12k(mg/k)2= 0.5(3 kg×9.8 m/s2)2÷
(240 N/m) = 1.80 J. (b) The change in just the gravitational potential energy between the equilibrium and unstretched positions is ∆UBA
g = Ug(A) −
Ug(B) = mgx = mg(mg/k) = 3.60 J. (c) The
corresponding change in the spring’s potential energy is ∆UBA
s = Us(A) − Us(B) = 0 − 1.80 J. Although the
change in the total potential energy, ∆UBA=
∆UBA
g + ∆UsBA= 3.60 J − 1.80 J = 1.80 J, is not
zero, there is no discrepancy. In order to move the fish slowly upward, you would have to exert an upward applied force (Fa = mg − kx) that would do work
WBA
a = mgx −12kx2= ∆UBA, as required by the
work-energy theorem.
Problem
19. The force exerted by a rubber band is given approximately by F = F0 ℓ + x ℓ − ℓ2 (ℓ + x)2 , where ℓ is the unstretched length, and F0 is a
constant. Find the potential energy of the rubber band as a function of the distance x it is stretched. Take the zero of potential energy in the
unstretched position.
Solution
Assuming that the direction of F is opposite to the displacement, we find: U = − Z x 0 F · dx ′=F 0 Z x 0 ℓ + x′ ℓ − ℓ2 (ℓ + x′)2 dx′ = F0 x′+x ′2 2ℓ + ℓ2 (ℓ + x′) x 0
Problem 18 Solution. = F0 x +x 2 2ℓ + ℓ2 (ℓ + x)− ℓ .
(This problem is identical to Problem 7-33. x′ is a
dummy variable.)
Section 8-3:
Conservation of Mechanical
Energy
Problem
20. A skier starts down a frictionless 32◦slope. After
a vertical drop of 25 m, the slope temporarily levels out, then drops at 20◦an additional 38 m
vertically before leveling out again. What is the skier’s speed on the two level stretches?
Solution
Since the slope is frictionless, mechanical energy is conserved. Thus, ∆KAB = −∆UAB, or 12mvB2 =
mg(yA− yB). Therefore, vB=
q
2(9.8 m/s2)(25 m) = 22.1 m/s. Repeating for ∆KAC= −∆UAC, we find
vC =
q
2(9.8 m/s2)(63 m) = 35.1 m/s. (We assumed that the skier started from rest, vA= 0. Note that the
exact profile of the slope is irrelevant.)
Problem 20 Solution.
Problem
21. A Navy jet of mass 10,000 kg lands on an aircraft carrier and snags a cable to slow it down. The cable is attached to a spring with spring constant 40,000 N/m. If the spring stretches 25 m to stop the plane, what was the landing speed of the plane?
Solution
If we assume no change in gravitational potential energy, Kinitial=12mv2= Ufinal= 12kx2, or v =
pk/m x. Thus, v = p(40,000 N/m)/(10,000 kg)× (25 m) = 50 m/s.
Problem
22. A spring of constant k, compressed a distance x, is used to launch a mass m up a frictionless slope that makes an angle θ with the horizontal. Find an expression for the maximum distance along the slope that the mass moves after leaving the spring.
Solution
If the slope is frictionless (and there are no other losses of energy), the total mechanical energy of the spring and mass is conserved. Initially, K0= 0, and
U0=12kx2+ mgy0 (where we neglect the gravitational
potential energy of the spring). Then Equation 8-8 gives U + K = 1
2kx
2+ mgy
0. After the mass has left
the spring, the potential energy is just the gravitational potential energy of the mass mgy, so mg(y − y0) = 12kx2− K. In terms of the distance, s,
along the slope, y − y0= (s − s0) sin θ. The maximum
distance occurs when the kinetic energy is zero, or (s − s0)max= kx2/2 mg sin θ.
Problem
23. A 120-g arrow is shot vertically from a bow whose effective spring constant is 430 N/m. If the bow is drawn 71 cm before shooting the arrow, to what height does the arrow rise?
Solution
If we ignore losses in energy due to air resistance etc., the mechanical energy of the bow and arrow is the
same just before shooting the arrow and when the arrow is at its maximum height, U0+ K0= U + K.
Before shooting, K0= 0, and U0= 12kx2+ mgy0, the
potential energy of the taut bow plus the gravitational energy of the arrow at the initial position. At the maximum height, K = 0 (instantaneously) and U = mgymax. Therefore, 12kx2+ mgy0= mgymax, or
ymax− y0= kx2/2mg = (430 N/m)(0.71 m)2÷
2(0.12 kg)(9.8 m/s2) = 92.2 m. (It is assumed that any change in the gravitational potential energy of the bow is negligible.)
Problem
24. A child is on a swing whose 3.2-m-long chains make a maximum angle of 50◦ with the vertical.
What is the child’s maximum speed?
Solution
This is precisely the situation described in Example 8-6(a). Thus, v =p2gℓ(1 − cos θ0) =
q
2(9.8 m/s2)(3.2 m)(1 − cos 50◦) = 4.73 m/s.
Problem
25. With x − x0= h and a = g, Equation 2-11 gives
the speed of an object thrown downward with initial speed v0 after it has dropped a distance
h: v =pv2
0+ 2gh. Use conservation of energy to
derive the same result.
Solution
Gravity is a conservative force, so for free fall near the Earth’s surface, U0+ K0= U + K, or 12mv02+ mgy0=
1 2mv
2+ mgy. (We neglect the effects of air resistance.)
Here, the y-axis is positive upward, so that the gravitational potential energy, relative to zero potential energy at y = 0, has the usual form mgy. Then the distance dropped is y0− y = h, instead of
x − x0= h, and ay= −g. Canceling m, and
rearranging terms, one recaptures the stated result.
Problem
26. In a switchyard, freight cars start from rest and roll down a 2.8-m incline and come to rest against a spring bumper at the end of the track
(Fig. 8-27). If the spring constant is 4.3×106N/m,
how much is the spring compressed when hit by a 57,000-kg freight car?
Solution
The freight car starts and ends at rest (no kinetic energy), so if we neglect friction, the work-energy theorem requires Wnet= ∆K = 0 = Wg+ Ws. The
2.8 m
figure 8-27 Problem 26.
work done by gravity is Wg = mg(2.8 m), and the
work done by the spring force is Ws= −12kx2
(the spring force is opposite to the displacement). Therefore, x =p2Wg/k =
q
2(5.7×104kg)(9.8 m/s2)(2.8 m)/(4.3×106N/m) =
85.3 cm. (Of course, for conservative forces, Wg+ Ws= −∆(Ug+ Us), and the work-energy
theorem is just the conservation of mechanical energy.)
Problem
27. Two clever kids use a huge spring with k = 890 N/m to launch their toboggan at the top of a 9.5-m-high hill (Fig. 8-28). The mass of kids plus toboggan is 80 kg. If the kids manage to compress the spring 2.6 m, (a) what will be their speed at the bottom of the hill? (b) What fraction of their final kinetic energy was initially stored in the spring? Neglect friction.
figure 8-28 Problem 27.
Solution
If friction is neglected everywhere (in the spring, on the snow, through the air, etc.), the mechanical energy of the kids plus toboggan (including potential energy of gravitation and the spring, as well as kinetic energy) is conserved: Utop+ Ktop= Ubot+ Kbot. At the top of
the hill, Ktop= 0 (the toboggan starts from rest), and
Utop= Utop,g+ Utop,s= mgytop+12kx2, while at the
bottom, Kbot=12mv2 and Ubot= mgybot(since
Ubot,s= 0 when the spring is no longer compressed).
(b) Using the given data, we find Kbot= mg(ytop−
ybot) +12kx2= (80 kg)(9.8 m/s2)(9.5 m)+ 1
2(890 N/m)(2.6 m)
Utop,s/Kbot= 3.01 kJ/10.5 kJ = 28.8%. (a) The
speed at the bottom is vbot=p2Kbot/m =
p2(10.5 kJ)/(80 kg) = 16.2 m/s.
Problem
28. A 200-g block slides back and forth on a friction-less surface between two springs, as shown in Fig. 8-29. The left-hand spring has k = 130 N/m, and its maximum compression is 16 cm. The right-hand spring has k = 280 N/m. Find (a) the maximum compression of the right-hand spring and (b) the speed of the block as it moves between the springs.
figure 8-29 Problem 28.
Solution
In the absence of friction and air resistance, the mechanical energy of the block is conserved. If the surface is horizontal (no change in Ugrav), then
1
2kℓx2ℓ= 12krx2r, where xℓ and xrare the maximum
compressions of the left- and right-hand springs, respectively (at which points the block is
instantaneously at rest). Thus, xr= xℓpkℓ/kr =
(16 cm)p130/280 = 10.9 cm. (b) At some point between the springs, 1
2mv 2=1 2kℓx 2 ℓ, or v = xℓpkℓ/m =p(130 N/m)/(0.2 kg)(0.16 m) =
4.08 m/s. (Of course, consideration of the right-hand spring must give the same result.)
Problem
29. An initial speed of 2.4 km/s (the “escape speed”) is required for an object launched from the moon to get arbitrarily far from the moon. At a mining operation on the moon, 1000-kg packets of ore are to be launched to a smelting plant in orbit around the Earth. If they are launched with a large spring whose maximum compression is 15 m, what should be the spring constant of the spring?
Solution
For an ideal spring (without losses), 12ky2= 1 2mv2B+
mgy. (This is Equation 8-7 for points A and B in the sketch.) Since the gravitational potential energy change is negligible compared to the other terms, k ≃ m(vB/y)2= 103kg(2.4 km/s = 15 m)2= 25.6 MN/m.
(Note: The surface gravity on the moon is 1.62 m/s2, so the maximum change in potential energy of a packet is only mgy = 24.3 kJ, while its kinetic energy,
1 2mv
2
B = 2.88 GJ, is more than 105 times larger.
Likewise, the gravitational potential energy of the spring is negligible.)
Problem 29 Solution.
Problem
30. A runaway truck lane heads uphill at 30◦ to the
horizontal. If a 16,000-kg truck goes out of control and enters the lane going 110 km/h, how far along the ramp does it go? Neglect friction.
Solution
If we neglect losses in mechanical energy, −∆K =
1 2mv
2
A= ∆U = mgy = mgℓ sin 30◦. Therefore, ℓ =
v2 A/g = (110 m/3.6 s)2/(9.8 m/s 2 ) = 95.3 m. Problem 30 Solution.
Problem
31. A low-damage bumper on a 1400-kg car is mounted on springs whose total effective spring constant is 7.0×105N/m. The springs can undergo
a maximum compression of 16 cm without damage to the bumper, springs, or car. What is the maximum speed at which the car can collide with a stationary object without sustaining damage?
Solution
The maximum energy stored by the springs, 12kx2,
without incurring damage, is equal to the maximum kinetic energy, 12mv2, of the car before a collision. (We
assume a level road and horizontal collision so that there is no change in the car’s gravitational potential energy.) Thus, 12kx2=1
2mv2, or v =pk/m x =
p(7.0×105N/m)/(1400 kg)(0.16 m) = 3.58 m/s =
12.9 km/h.
Problem
32. A block slides on the frictionless loop-the-loop track shown in Fig. 8-30. What is the minimum height h at which it can start from rest and still make it around the loop?
Solution
From the conservation of mechanical energy (no friction) KA+ UA= KB+ UB, mgyA= 12mv2B+ mgyB,
or v2
B= 2g(h − 2R). The condition that the block
stay on the track is v2
B≥ gR (see the solution to
Problem 68 below), so 2g(h − 2R) ≥ gR or h ≥ 5R/2.
figure8-30 Problem 32 Solution.
Problem
33. Show that the rope in Example 8-6 can remain taut all the way to the top of its smaller loop only if a ≤ 25ℓ. (Note that the maximum release angle is
90◦for the rope to be taut on the way down.)
Solution
For the rope to be taut at the top of the small circle, the tension must be greater than (or equal to) zero; T = mv2
top/a − mg ≥ 0, or v2top≥ ga. Therefore, the
mechanical energy at the top is greater than (or equal to) a corresponding value: E = K + U = 12mv2top+
mg(2a) ≥ 1
2m(ga) + 2mga = 5
2mga, where the zero
of potential energy is the lowest point (as in Example 8-6). Since energy is conserved, E = U0=
mgℓ(1 − cos θ0), where θ0 is the release angle (see
Example 8-6 again), so a ≤ 2E/5mg = 2
5ℓ(1 − cos θ0).
The greatest upper limit for the radius a corresponds
to the maximum release angle, as stated in the question above, since cos 90◦= 0.
Problem
34. The maximum speed of the pendulum bob in a grandfather clock is 0.55 m/s. If the pendulum makes a maximum angle of 8.0◦with the vertical,
what is the length of the pendulum?
Solution
From Example 8-6(a), v2= 2gℓ(1 − cos θ
0), or ℓ =
v2/2g(1 − cos θ
0) = (0.55 m/s)2/2(9.8 m/s2)×
(1 − cos 8◦) = 1.59 m.
Problem
35. A 2.0-kg mass rests on a frictionless table and is connected over a frictionless pulley to a 4.0-kg mass, as shown in Fig. 8-31. Use conservation of energy to calculate the speed of the masses after they have moved 50 cm.
2 kg
4 kg
figure 8-31 Problem 35.
Solution
Since friction is absent by assumption, mechanical energy is conserved, and U0+ K0= U1+ K1, where
the subscripts 0 and 1 refer to the initial position (both masses at rest) and the final position (after each has moved 50 cm), respectively. Then U0− U1= K1=
(4 kg)(9.8 m/s2)(0.5 m) =12(2 kg + 4 kg)v2, since the
potential energy changes only for the falling mass, and both masses have the same speed. Solving for v, we find v = 2.56 m/s. (The kinetic energy of the pulley was assumed to be negligible.)
Problem
36. The masses shown in Fig. 8-32 are connected by a massless string over a frictionless, massless pulley and are released from rest. Use energy
conserva-tion to find (a) the velocity of the 7.0-kg mass just before it hits the floor, (b) the maximum height reached by the 4.0-kg mass, and (c) the fraction of the system’s initial mechanical energy lost when the 7.0-kg mass comes to rest on the floor.
figure 8-32 Problem 36.
Solution
If we ignore possible losses (friction and air resistance) and the energy of the string and pulley (both assumed massless), then the total mechanical energy of the masses is conserved. Each mass has gravitational potential energy mgy, measured from zero on the floor, and kinetic energy 12mv2. Since the masses are
connected (by a string of presumably fixed length), their velocities have equal magnitudes when both masses are off the floor. (a) If we take position A when the 4-kg mass is on the floor at rest, and position B when the 7-kg mass is about to strike the floor, then the conservation of energy requires UA+ KA= UB+ KB, or (7×9.8 N)(5 m) = 1 2(7 kg)v 2 B+12(4 kg)v 2 B+ (4×9.8 N)(5 m). Thus, vB =p2(3×9.8 N)(5 m)/(11 kg) = 5.17 m/s.
(b) The 4-kg mass, with an upward velocity of 5.17 m/s, will rise (as a projectile) an additional v2
B/2g = (5.17 m/s)2/(2×9.8 m/s 2
) = 1.36 m to a maximum height of 5 m + 1.36 m = 6.36 m off the floor. (Note: the mechanical energy of both masses is no longer conserved after the 7-kg mass hits the floor, because a non-conservative contact force acts to stop it.) (c) The kinetic energy of the 7-kg mass is lost when it strikes the floor. 12(7 kg)(5.17 m/s)2= 93.5 J
is about 27.3% of the initial energy of the system, which was (7×9.8 N)(5 m) = 343 J.
Problem
37. A mass m is dropped from a height h above the top of a spring of constant k that is mounted vertically on the floor (Fig. 8-33). Show that the maximum compression of the spring is given by (mg/k)(1 +p1 + 2kh/mg). What is the significance of the other root of the quadratic equation?
Solution
If the maximum compression is y, as shown, and we measure gravitational potential energy from the lowest point, B, then the conservation of energy between point A and B requires that mg(h + y) = 1
2ky 2. The
quadratic formula can be used to find y = (mg/k)× (1 ±p1 + 2kh/mg). Only positive values of y are physically meaningful in this problem, because the spring is not compressed unless y > 0.
figure 8-33 Problem 37 Solution.
Section 8-4:
Potential Energy Curves
Problem
38. A particle slides along the frictionless track shown in Fig. 8-34, starting at rest from point A. Find (a) its speed at B, (b) its speed at C, and (c) the approximate location of its right-hand turning point.
Solution
The particle has kinetic energy 12mv2and
gravitational potential energy mgy (measured above the reference level y = 0 in Fig. 8-34), so in the absence of friction, the sum of these is a constant. We are given that vA= 0 and yA= 3.8 m, so we can
evaluate the constant and express the mechanical energy at any other point in terms of it:
D y
x
figure 8-34 Problem 38.
1
2mv2+ mgy = mgyA. (a) Solving for the speed at
point B, we find vB=p2g(yA− yB) =
q
2(9.8 m/s2)(3.8 m − 2.6 m) = 4.85 m/s. (b) Similarly, vC= 7.00 m/s. (c) The right-hand
turning point is the point D where the particle’s velocity is instantaneously zero, before changing direction back to the left in Fig. 8-34. Thus, yA= yD,
and from the figure, we estimate that xD is about
11 m.
Problem
39. A particle slides back and forth on a frictionless track whose height as a function of horizontal position x is given by y = ax2, where a =
0.92 m−1. If the particle’s maximum speed is
8.5 m/s, find the turning points of its motion.
Solution
With no friction, the only forces acting are gravity, a conservative force with potential energy U = mgy above a reference level at y = 0, and the normal contact force of the track. Although the latter is nonconservative, it is always perpendicular to the displacement along the track (by definition); hence it does no work on the particle, and the mechanical energy is conserved: 12mv2+ mgy = constant. Since
the height y = ax2 is given in terms of the horizontal
displacement, we may write 12mv2+ mgax2=
constant, or v2+ 2gax2= (a different) constant. One
can see that the maximum speed occurs when the displacement is a minimum (zero, in fact) and vice-versa. Thus, one can determine the constant two ways: v2+ 2gax2= v2
max= 2gax2max. The turning
points are the places where the velocity is instantaneously zero (a minimum) and hence are given by xmax= ±pv2max/2ga =
± q
(8.5 m/s)2/2(9.8 m/s2)(0.92 m−1) = ±2.00 m.
(In this case, the particle’s motion is oscillatory, back and forth between the turning points, with maximum speed, either forward or backward, at the middle, x = 0.)
Problem
40. A particle with total energy 3.5 J is trapped in a potential well described by U = 7.0 − 8.0x + 1.7x2,
where U is in joules and x in meters. Find its turning points.
Solution
A particle in a one-dimensional potential well, which conserves mechanical energy, satisfies Equation 8-7 with total energy E = K + U =12m(dx/dt)2+ U (x).
The turning points are the solutions of this equation when dx/dt = 0; that is, E = U (x) = 3.5 = 7.0 − 8.0x + 1.7x2 (where energy is in joules and
displacement in meters). The quadratic formula gives x = (1.7)−1(4.0 ±√10.1) m = 0.488 m and 4.22 m.
Problem
41. The potential energy associated with a
conservative force is shown in Fig. 8-35. Consider particles with total energies E1= −1.5 J, E2=
−0.5 J, E3= 0.5 J, E4= 1.5 J, and E5= 3.0 J.
Discuss the subsequent motion, including the approximate location of any turning points, if the particles are initially at point x = 1 m and moving in the −x direction.
figure 8-35 Problem 41 Solution.
Solution
All the particles start in the left-hand potential well and reverse direction when they hit the left-most (infinite) potential barrier. The first four particles have insufficient energy to escape from this well (the height of the next barrier is about 2.0 J > E4) and
experience a second turning point (between x = 2.0 m and 4.0 m, depending on energy). The fifth particle’s motion is unbounded.
Problem
42. Make an accurate potential energy curve, covering the region −8 m < x < 8 m, for potential energy U = (ax2− b)e−x2
/c2
, where a = 1.5 J/m2, b = 5.0 J, and c = 3.0 m. Discuss the subsequent motion of 1-kg particles starting from the origin
and moving initially in the +x direction with total energies of −3 J, 1 J, and 4 J. Include the location of any turning points. Determine also the speed of the highest-energy particle when it is a great distance from the origin.
Solution
The potential energy is symmetric about x = 0× (U (−x) = U(x)), so we need only plot it for positive x. The zeros of U are at x = ±pb/a = ±1.83 m. The maxima and minimum can be found from the zeros of the derivative, dU/dx = (2x/c2)(b + ac2− ax2)e−x2
/c2 , which are xmin= 0 and xmax= ±pc2+ b/a =
±3.51 m. The corresponding energies are Umin= −b =
−5.0 J, and Umax= ac2e−(1+b/ac
2
)= 3.43 J. We
finally note that U (8 m) = 91 e−7.11 J = 7.43×10−2 J.
One can see that the particles with energies less than Umax are confined to the well, with turning points on
either side of ±1.83 m, respectively. The speed of a particle is v =p2(E − U)/m. Since U → 0 for x → ∞, the speed of the particle with E = 4 J approachesp2(4 J)/(1 kg) = 2.83 m/s for large x.
Problem 42 Solution.
Problem
43. (a) Derive an expression for the potential energy of an object subject to a force Fx= ax − bx3,
where a = 5 N/m, and b = 2 N/m3, taking U = 0 at x = 0. (b) Graph the potential energy curve for x > 0, and use it to find the turning points for an object whose total energy is −1 J.
Solution
(a) The force is conservative (any one-dimensional force given by an integrable function of position is), so the potential energy can be found from Equation 8-2a:
U (x2) − U(x1) = − Z x2 x1 (ax − bx3) dx = −a2(x22− x21) + b 4(x 4 2− x41).
If we define the zero of potential energy at x = 0, then U (x) = −1
2ax 2+1
4bx
4. (b) A graph of U (x) for x ≥ 0,
when a = 5 N/m, b = 2 N/m3 and x is in meters, is shown. (Note that the potential energy is symmetric, U (−x) = U(x), but that only positive displacements are considered in this problem.) The conservation of energy can be written in terms of the total energy, E =12m(dx/dt)2+ U (x), so that dx/dt =
±p2[E − U(x)]/m. The maximum speed occurs when U (x) is a minimum; i.e., dU/dx = 0, and d2U/dx2> 0.
Taking the derivative, one finds 0 = −ax + bx3, which
has solutions x = 0 and x = ±pa/b = ±p5/2 m = ±1.58 m. The second derivative d2U/dx2= −a + 3bx2
is negative for x = 0, which is a local maximum, but is positive for x = ±pa/b, which are minima with Umin= U (±pa/b) = −a2/4b = −(25/8) J = −3.13 J.
There is real physical motion (K ≥ 0) for total energy E ≥ Umin. The turning points (where dx/dt = 0) can
be found from the equation U (x) = E; there are four solutions (two positive) for energies with Umin< E <
0, and two solutions (one positive) for E > 0.
Problem 43 Solution.
The equation U (x) − E = 0 is equivalent to x4 −
2(a/b)x2− 4(E/b) = 0. The quadratic formula gives
x = ±{(a/b) ± [(a/b)2+ 4(E/b)]1/2}1/2 for U min<
E < 0, and x = ±{(a/b) + [(a/b)2+ 4(E/b)]1/2}1/2 for
E > 0. For the particular values given (E = −1 J), the positive turning points are x =
q
(5 ±√17)/2 m = 0.662 m and 2.14 m, as can be seen in the graph.
Section 8-5:
Force and Potential Energy
Note: In the following problems, motion is restricted to one dimension.
Problem
44. Figure 8-36 shows the potential energy curve for a certain particle. Find the force on the particle at each of the curve segments shown.
0 1 2 3 –1 –2 –3 U (x ) (J) 1 2 4 5 6 (a) (b) (c) (d) (f ) (e) x (m) figure 8-36 Problem 44.
Solution
The force on a particle in one-dimensional motion, when the potential energy is a straight line segment, is the negative of the slope of U (x); i.e., Fx= −dU/dx
(Equation 8-8). (This is, of course, just the conservative force represented by the potential energy.) From Fig. 8-36, we can determine the negative of the slopes: (a) −3 J/1.5 m = −2 N; (b) 0; (c) −(−4 J/1
2 m) = 8 N; (d) −(−1 J/1 m) = 1 N,
(e) −4 N/1 m = −4 N, (f) 0.
Problem
45. A particle is trapped in a potential well described by U (x) = 2.6x2− 4, where U is in joules, and x is
in meters. Find the force on the particle when it’s at (a) x = 2.1 m; (b) x = 0 m; and (c) x = −1.4 m.
Solution
For one-dimensional motion, Equation 8-8 gives the force Fx= −dU/dx = −d/dx(1.6x2− 4) = −2(1.6)x =
−3.2x, where Fx is in newtons for x in meters.
Therefore, (a) Fx(2.1 m) = −(3.2)(2.1) N = −6.72 N
(force in the negative x direction); (b) Fx(0) = 0;
(c) Fx(−1.4 m) = −(3.2)(−1.4) N = 4.48 N.
Problem
46. In ionic solids such as NaCl (salt), the potential energy of a pair of ions takes the form
U = br−n− ar−1, where r is the separation of the
ions. For NaCl, a and b have the SI values 4.04×10−28 and 5.52×10−98, respectively, and
n = 8.22. Find the equilibrium separation between ions in NaCl.
Solution
At the equilibrium separation, the potential energy is a minimum, or (dU/dr)eq= 0 = −nbreq−(n+1)+ areq−2.
Then req= (nb/a)1/(n−1)= (8.22×5.52×10−98÷
4.04×10−28)1/7.22 (SI units) = 2.82×10−10m =
2.82 ˚A. (The angstrom is a common non-SI unit of length, used in chemistry and atomic physics; see Appendix C.)
Problem
47. The potential energy associated with a certain conservative force is given by U = bx2, where b is
a constant. Show that the force always tends to accelerate a particle toward the origin if b is positive and away from the origin if b is negative.
Solution
The conservative force represented by the
one-dimensional potential energy U (x) = bx2 is given
by Equation 8-8, Fx= −dU/dx = −2bx. x is the
displacement from the origin, so this is towards the origin for b > 0 and away from the origin for b < 0. (The former is called a restoring force.)
Problem
48. A more accurate expression for the potential well in Fig. 8-18 than that provided in Example 8-7 is U = 286(x − xe)2− 6.22×1012(x − xe)3 with U in
joules and x in meters. Find the force on the hydrogen atoms when they are 0.10 nm apart.
Solution
The force along the line joining the hydrogen atoms is Fx= −dU/dx = −2(286)(x − xe) + 3(6.22×1012)×
(x − xe)2. For x − xe= (0.1 − 0.0741) nm = 2.59×
10−11 m, the force is F
x= (−1.48 + 1.25)×10−8N =
−2.32 nN. (The direction of the force is toward the equilibrium separation, xe.)
Problem
49. The potential energy of a spring is given by U = ax2− bx + c, where a = 5.20 N/m, b = 3.12 N, and c = 0.468 J, and where x is the overall length of the spring (not the stretch). Find (a) the
equilibrium length of the spring and (b) the spring constant.
Solution
(a) The natural, or unstretched, length of the spring is the value of x for which the spring force is zero. Thus, Fx= −dU/dx = −2ax + b = 0, when x = b/2a =
(3.12 N)/2(5.2 N/m) = 30 cm. (b) Hooke’s Law defines the spring constant in terms of the stretch (in this case x − b/2a). Since Fx= −2a(x − b/2a), the
spring constant is k = 2a = 10.4 N/m. (Alternatively, k = −dFx/dx = d2U/dx2.)
Section 8-6:
Nonconservative Forces
Problem
50. Repeat Problem 20 for the case when the
coefficient of kinetic friction on both slopes is 0.11, while the level stretches remain frictionless.
Solution
The work done by friction skiing down a straight slope of length ℓ is Wf = −fkℓ = −µkN ℓ = −µk(mg cos θ)×
(h/ sin θ) = −µkmgh cot θ, where h = ℓ sin θ is the
ver-tical drop of the slope. The energy principle applied between the start and the first level (see Problem 20) now gives ∆KAB+ ∆UAB = Wf,AB, or 12mvB2 =
mg(yA− yB) − µkmg(yA− yB) cot 32◦. Therefore,
vB =
q
2(9.8 m/s2)(25 m)(1 − 0.11 cot 32◦) =
20.1 m/s. When we repeat this argument for the motion between the top and the second level, we must include all the work done by friction, so
∆KAC+ ∆UAC= Wf,AB+ Wf,BC. Therefore, vC=
p
2(9.8 m/s2
)[63 m − (0.11)(25 m) cot 32◦ − (0.11)(38 m) cot 20◦]
= 30.4 m/s.
(The alternate expression, ∆KBC+ ∆UBC= Wf,BC,
gives the same result,
vC=
p
2(9.8 m/s2)(38 m)(1 − 0.11 cot 20◦) + (20.1 m/s)2.)
Problem
51. A basketball dropped from a height of 2.40 m rebounds to a maximum height of 1.55 m. What fraction of the ball’s initial energy is lost to nonconservative forces? Take the zero of potential energy at the floor.
Solution
The energy principle in the form of Equation 8-5 is Wnc= ∆K + ∆U. Since v = 0 where the ball is
dropped (point A) and at the maximum of its rebound (point B), one has Wnc= ∆UAB= mg(yB− yA). This
is a loss of energy since yB < yA. The initial energy was
just mgyA, relative to zero of potential energy on the
floor at y = 0, so the fraction lost is mg |yB− yA| ÷
mgyA= 1 − yB/yA= 1 − 1.55/2.40 = 35.4%. (The
words “loss” and “lost” replace the minus sign in the energy change; i.e., a “35% loss” equals a change of −35%.)
Problem
52. A 1.5-kg block is launched up a 30◦ incline with
an initial speed of 6.4 m/s. It comes to a halt after moving 3.4 m along the incline, as shown in Fig. 8-37. Find (a) the change in the block’s kinetic energy; (b) the change in the block’s potential energy; (c) the work done by friction; (d) the coefficient of kinetic friction.
figure 8-37 Problem 52.
Solution
Let the initial and final positions of the block in Fig. 8-37 be denoted by points A and B, (a) ∆KAB= 1 2m(v 2 B− vA2) =12(1.5 kg)(0 − (6.4 m/s) 2) = −30.7 J. (b) ∆UAB = mg(yB− yA) = (1.5 kg)(9.8 m/s2)×
(3.4 m) sin 30◦= 25.0 J. (c) From Equation 8-5,
Wf,AB= ∆KAB+ ∆UAB = −5.73 J. (d) The force of
friction on the incline is fK= µKN = µKmg cos 30◦
opposite to the direction of motion, so the work done by friction is Wf,AB = −fkℓ = −µKmgℓ cos 30◦. Thus
µK= −Wf,AB/mgℓ cos 30◦= 5.73 J/(1.5 kg)×
(9.8 m/s2)(3.4 m) cos 30◦= 0.13.
Problem
53. A pumped-storage reservoir sits 140 m above its generating station and holds 8.5×109 kg of water.
The power plant generates 330 MW of electric power while draining the reservoir over an 8.0-hour period. What fraction of the initial potential energy is lost to nonconservative forces (i.e., does not emerge as electricity)?
Solution
If all the water fell through the same difference in height, the amount of gravitational potential energy released would be ∆U = mg∆y = (8.5×109kg)×
(9.8 m/s2)(140 m) = 1.17×1013J. The energy
generated by the power plant at an average power output of 330 MW over an 8 h period is (330 MW)× (8×3600 s) = 9.50×1012 J, so the fraction lost is
Problem
54. A spring of constant k = 340 N/m is used to launch a 1.5-kg block along a horizontal surface whose coefficient of sliding fraction is 0.27. If the spring is compressed 18 cm, how far does the block slide?
Solution
Suppose the block comes to rest at B, a distance ℓ from its initial position at rest against the compressed spring at A. Applying Equation 8-5, we find Wnc=
−µkmgℓ = ∆K + ∆U = −UsA= −12kx
2, since the
kinetic energies at A and B and the change in gravitational potential energy are zero. Therefore, ℓ = 12kx2/µ
kmg = 0.5(340 N/m)(0.18 m)2/(0.27)×
(1.5 kg)(9.8 m/s2) = 1.39 m.
Problem 54 Solution.
Problem
55. A 2.5-kg block strikes a horizontal spring at a speed of 1.8 m/s, as shown in Fig. 8-38. The spring constant is 100 N/m. If the maximum compression of the spring is 21 cm, what is the coefficient of friction between the block and the surface on which it is sliding?
1.8 m/s
figure 8-38 Problem 55.
Solution
The final kinetic energy, the initial potential energy of the spring, and the change in the gravitational potential energy are all zero. Therefore, Wnc=
−µKmgx = ∆K + ∆U = −12mv20+12kx 2=
0.5[(100 N/m)(0.21 m)2−(2.5 kg)(1.8 m/s)2] = −1.85 J,
or µk = (1.85 J)/(2.5 kg)(9.8 m/s2)(0.21 m) = 0.36.
Problem
56. A meteorite strikes Earth and embeds itself 1.7 m into the ground. Scientists dig up the meteorite
and find that its mass is 400 g; they estimate that the ground exerted a retarding force of 106N on
the meteorite. Estimate the impact speed of the meteorite.
Solution
The average 106 N non-conservative frictional
retarding force acts over a distance of 1.7 m opposite to the direction of penetration in stopping the
meteorite. If we neglect the (small) change in potential energy during the penetration the work-energy
theorem gives Wnc= −(106N)(1.7 m) = ∆K + ∆U ≈
∆K = 0 −1 2mv 2 i = −12(0.4 kg)v 2 i, or vi= 2.92 km/s.
Problem
57. A surface is frictionless except for a region between x = 1 m, and x = 2 m, where the coefficient of friction is given by µ = ax2+ bx + c, with
a = −2 m−2, b = 6 m−1, and c = −4. A block is
sliding in the −x direction when it encounters this region. What is the minimum speed it must have to get all the way across the region?
Solution
Assume that the surface is horizontal, so that there are no changes in the block’s potential energy (∆U = 0) and the force of friction on it is fk= −µkN = −µkmg
(opposite to the direction of motion along the x-axis). If the block crosses the entire region from x1= 1 m to
x2= 2 m, the work-energy theorem demands that
Wnc= − Z x2 x1 µkmg dx = ∆K = 1 2m(v 2 2− v21) or v21− v22= 2g Z x2 x1 (ax2+ bx + c)dx = 2g a 3x 3+b 2x 2+ cx 2 m 1 m = 6.53 m2/s2. (The given values of a, b, and c were used.) The minimum speed at the start of the region is the value of v1 when v2= 0, or v1,min=p6.53 m2/s2 =
2.56 m/s.
Problem
58. A biologist uses a spring-loaded dart gun to shoot a 50-g tranquilizing dart into an elephant 21 m away. The gun’s spring has spring constant k = 690 N/m and is pulled back 14 cm to launch the dart. The dart embeds itself 2.2 cm in the elephant. (a) What is the average stopping force exerted on the dart by the elephant’s flesh? (b) How long does it take the dart to reach the
elephant? Assume the dart’s trajectory is nearly horizontal.
Solution
(a) Assume that there is no change in the
gravitational potential energy of the dart, and that the stopping force is the only non-negligible, non-conservative force. Then Wnc= −F (2.2 cm) =
∆U + ∆K = −1 2kx
2= −1
2(690 N/m)(0.14 m) 2, or
F = 307 N. (b) The (horizontal) speed of the dart as it leaves the spring is v =pkx2/m =
p(690 N/m)(0.14 m)2/(0.05 kg) = 16.4 m/s, so the
time it takes to travel 21 m (horizontally) is approximately t = d/v = 21 m/(16.4 m/s) = 1.28 s.
Problem
59. A skier starts from rest at the top of the left-hand peak in Fig. 8-39. What is the maximum
coefficient of kinetic friction on the slopes that would allow the skier to coast to the second peak? (Your answer, of course, neglects air resistance.)
figure8-39 Problem 59 Solution.
Solution
Assume that the maximal frictional force of the slopes and gravity are the only significant forces acting on the skier, who starts from rest on the higher peak (point A) and finishes at rest just reaching the lower peak (point B). Then the work-energy theorem requires that Wf = ∆U = mg(yB− yA). The force of
friction on each slope is fk= −µmaxk mg cos θ, and the
displacement along each slope is ℓ = y/ sin θ (measured from the bottom of both slopes), so that work done by friction on each slope is fkℓ = (−µmaxk mg cos θ)×
(y/ sin θ) = −µmax
k mgy cot θ. Then the work-energy
theorem becomes −mg(yA− yB) = −µmaxk mg×
(yAcot θA+ yBcot θB), from which µmaxk can be
determined: µmax
k = (950 − 840)/(950 cot 27◦+
840 cot 35◦) = 0.036.
Problem
60. A bug slides back and forth in a hemispherical bowl of 11 cm radius, starting from rest at the top, as shown in Fig. 8-40. The bowl is frictionless except for a 1.5-cm-wide sticky patch at the bottom, where the coefficient of friction is 0.61.
How many times does the bug cross the sticky region?
11 cm
1.5 cm
figure 8-40 Problem 60.
Solution
Each time it crosses the sticky patch, the bug loses energy ∆E = Wnc= −µkmg (1.5 cm). The initial
energy of the bug is E0= mg(11 cm) (measured above
the bottom). Since E0/ |∆E| = (11 cm)/(0.61)×
(1.5 cm) = 12.02, the bug will just barely complete twelve crossings.
Problem
61. A 190-g block is launched by compressing a spring of constant k = 200 N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction µ = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in Fig. 8-41. After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.
Solution
The energy of the block when it first encounters friction (at point O) is K0=12(200 N/m)(0.15 m)2=
2.25 J, if we take the zero of gravitational potential energy at that level. Crossing the frictional zone, the block loses energy ∆E = Wnc= −µkmgℓ = −(0.27)×
(0.19 kg)(9.8 m/s2)(0.85 m) = −0.427 J. Since K0/ |∆E| = 5.27, five complete crossings are made,
leaving the block with energy K0− 5 |∆E| = 0.113 J
on the curved rise side. This remaining energy is sufficient to move the block a distance s = 0.113 J ÷ µkmg = 22.5 cm towards point O, so the block comes
to rest 85 − 22.5 = 62.5 cm to the right of point O.
Section 8-7:
Conservation of Energy and
Mass-Energy
Problem
62. Two deuterium nuclei fuse to form a helium nucleus. Each deuterium has mass 3.344×
figure8-41 Problem 61 Solution.
10−27 kg, and the helium has mass
6.645×10−27 kg. Find the energy released in this
reaction.
Solution
The difference in mass between two deuterium nuclei and one helium nucleus is converted to energy, according to Einstein’s equation: E = ∆mc2=
(2×3.344 − 6.645)×10−27 kg×(3×108m/s)2=
3.87×10−12 J.
Problem
63. A hypothetical power plant converts matter entirely into electrical energy. Each year, a worker at the plant buys a box of 1-g raisins, and each day drops one raisin into the plant’s energy conversion unit. Estimate the average power output of the plant, and compare with a 500-MW coal-burning plant that consumes a 100-car trainload of coal every 3 days.
Solution
The energy equivalent of 1 g of mass is mc2, so
the power generated in one day is (10−3kg)×
(3×108m/s)2/(86,400 s) = 1.04 GW, or approximately
twice the output of a 500-MW coal-burning plant. Thus, the mass-energy of one raisin is equivalent to the chemical energy (times the overall efficiency) released by burning 67 car-loads of coal.
Problem
64. The Sun’s total power output is 3.85×1026W.
What is the associated rate at which the Sun loses mass?
Solution
With the use of Einstein’s equation, the power output of the Sun is found to be equivalent to a mass loss of dm/dt = d(E/c2)/dt = P/c2= (3.85×1026J/s) ÷
(3×108 m/s)2= 4.28×109 kg/s. (This is only about
7×10−14 solar masses per year.)
Paired Problems
Problem
65. A block slides down a frictionless incline that terminates in a ramp pointing up at a 45◦ angle,
as shown in Fig. 8-42. Find an expression for the horizontal range x shown in the figure, as a function of the heights h1 and h2shown.
Solution
After leaving the ramp (at point 2), with speed v2 at
45◦to the horizontal, the block describes projectile
motion with a horizontal range of x = v2 2/g (see
Equation 4-10). Since the track is frictionless (and the normal force does no work), the mechanical energy of the block (kinetic plus gravitational potential) is conserved between point 2 and its start from rest at point 1. Then ∆K = 12mv2
2− 0 = −∆U =
mg(h1− h2), and x = 2(h1− h2).
figure 8-42 Problem 65 Solution.
Problem
66. A block of mass m is launched horizontally from a compressed spring on a frictionless track that turns upward at a 45◦angle, as shown in Fig. 8-43.
Find an expression for the horizontal range x shown in the figure, as a function of the distance d by which the spring is initially compressed, the spring constant k, and the height h of the ramp.
figure 8-43 Problem 66 Solution.
Solution
The comments made in the solution to the previous problem hold here also, so x = v2
2/g, where point 2 is
the end of the ramp. Now there is potential energy of the spring as well as gravity. If point 1 is taken to be
the block’s position at maximum compression of the spring, then ∆K = 12mv2
2− 0 = −∆U = 12kd2− mgh.
Thus x = (kd2/mg) − 2h.
Problem
67. A ball of mass m is being whirled around on a string of length R in a vertical circle; the string does no work on the ball. (a) Show from force considerations that the speed at the top of the circle must be at least√Rg if the string is to remain taut. (b) Show that, as long as the string remains taut, the speed at the bottom of the circle can be no more than√5 times the speed at the top.
Solution
(a) At the top of the circle, the forces acting on the mass are gravity and the string tension, both
downward and parallel to the centripetal acceleration. Thus T + mg = mv2
top/R. Since Ttop≥ 0 if the string
is taut, v2
top≥ gR. (See Example 6-8.) (b) The
mechanical energy of the mass is conserved, since the tension does no work (by assumption), gravity is conservative, and air resistance is ignored. Thus Utop+ Ktop= Ubot+ Kbot= 12mv2top+ mgytop=
1
2mvbot2 + mgybot, or vbot2 = vtop2 + 2g(ytop− ybot) =
v2
top+ 4gR (since ytop− ybot is the diameter of the
circle). The result of part (a) then leads to v2
bot≤ v2top+ 4v2top= 5vtop2 , equivalent to the assertion
in the problem.
Problem
68. An 840-kg roller-coaster car is launched from a giant spring of constant k = 31 kN/m into a frictionless loop-the-loop track of radius 6.2 m, as shown in Fig. 8-44. What is the minimum amount that the spring must be compressed if the car is to stay on the track?
Solution
If the car stays on the track, the radial component of its acceleration is v2/R, and the normal force is
greater than zero. Thus, N = mv2/R + mg cos θ ≥ 0,
or v2≥ −gR cos θ. Now − cos θ has its maximum value
at the top of the loop (θ = 180◦), so v2
B ≥ gR is the
condition for the car to stay on the track all the way around. In the absence of friction, the conservation of mechanical energy requires KA+ UA= KB+ UB, or
0 + 1 2kx 2+ mgy A= 12mv2B+ mgyB. Therefore, x2= (m/k)(v2 B+ 2g(yB− yA)) ≥ (m/k)(gR + 2g(2R)) = 5mgR/k, or x ≥ [5(840 kg)(9.8 m/s2)(6.2 m) ÷ (31,000 N/m)]1/2= 2.87 m.
figure 8-44 Problem 68 Solution.
Problem
69. A pendulum consisting of a mass m on a string of length ℓ is pulled back so the string is horizontal, as shown in Fig. 8-45. The pendulum is then released. Find (a) the speed of the mass and (b) the magnitude of string tension when the string makes a 45◦ angle with the horizontal.
figure 8-45 Problem 69 Solution.
Solution
(a) We assume that the mechanical energy of the pendulum mass is conserved (neglect possible losses), and that this consists of kinetic and gravitational potential energy. At point 1 where the mass is released, U1= mgy1 and K1= 0, while after a 45◦
swing to point 2, U2= mgy2, and K2=12mv22. Energy
conservation implies K2= U1− U2, or v22 =
2g(y1− y2) = 2gℓ sin 45◦=
√
2gℓ, where we used the trigonometry apparent in Fig. 8-45 to express the difference in height in terms of the length of the string and angle of swing. (b) The string tension is in the direction of the centripetal acceleration (also shown in Fig. 8-45 at point 2), so the radial component of Newton’s second law gives T2− mg sin 45◦= mv22/ℓ.
mg sin 45◦= (m√2gℓ/ℓ) + (mg/√2) = 3mg/√2.
Problem
70. A particle of mass m slides down a frictionless quarter-circular track of radius R as shown in Fig. 8-46. If it starts from rest at the top of the track, find (a) its speed and (b) the magnitude of the normal force exerted by the track when the radius vector from the center of the track to the particle makes a 45◦ angle, as shown.
figure 8-46 Problem 70.
Solution
The forces, initial conditions, and trajectory of the particle in this problem are completely analogous to those in the previous problem, as can be seen from a comparison of Figures 8-45 and 46. The answers are also analogous, with N replacing T and R replacing ℓ.
Problem
71. A particle slides back and forth in a frictionless bowl whose height is given by h(x) = 0.18x2,
where x and h are both in meters. If the particle’s maximum speed is 47 cm/s, find the x coordinates of its turning points.
Solution
This situation is exactly similar to that in Problem 39, where the energy principle led to12mv2max = mghmax=
mg(0.18/m)x2
max. Thus, the turning points occur at
xmax= ±
q
(0.47 m/s)2/2(9.8 m/s2)(0.18/m) =
±25.0 cm.
Problem
72. A 1-kg particle slides on a frictionless track whose height is given by y = ax4− bx2, where
a = 1 m−3, and b = 4 m−1. The particle’s total
energy is 20 J, where the zero of potential energy is at y = 0. Find the turning points of its motion.
Solution
At the turning points, the kinetic energy is
(instantaneously) zero, so the energy principle (with gravitational potential energy U zero at y = 0 and Wnc= 0 for the normal force of the track) gives
E = 20 J = mgymax= mg(ax4max− bx2max). The
quadratic formula (positive solution for x2max) yields
x2
max= (b +pb2+ 4aE/mg)/2a = 4.46 m2, so
xmax= ±2.11 m.
Problem
73. A child sleds down a frictionless hill whose vertical drop is 7.2 m. At the bottom is a level but rough stretch where the coefficient of kinetic friction is 0.51. How far does she slide across the level stretch?
Solution
The child starts near the hilltop with KA= 0 and
stops on rough level ground, KB= 0, after falling
through a potential energy difference ∆U = UB−
UA= −mg(yA− yB), where yA− yB= 7.2 m. The
work done by friction (on level ground, N = mg) is Wnc= −fkx = −µkmgx, where x is the distance slid
across the rough level stretch. The energy principle, Equation 8-5, relates these quantities: Wnc=
−µkmgx = ∆K + ∆U = 0 − mg(yA− yB). Thus
x = (yA− yB)/µk = 7.2 m/0.51 = 14.1 m.
Problem
74. At the bottom of a frictionless ski slope is a 20-m-wide stretch of rough snow where the coefficient of friction is 0.32. From what vertical height on the slope should a skier start at rest, in order to get through the rough patch and come out of it with a speed of 6.0 m/s?
Solution
This situation is similar to the previous problem (if the rough stretch is still level) but with KB =12mvB2
and vB= 6.0 m/s. Thus Wnc= −µkmgx = 12mv2B− mg(yA− yB), or yA− yB= (vB2/2g) + µkx = (6 m/s)2/(19.6 m/s2 ) + (0.32)(20 m) = 8.24 m.
Supplementary Problems
Problem
75. A uranium nucleus has a radius of 1.43×10−10 m.
surface of the nucleus with negligible speed, subject to a repulsive force whose magnitude is F = A/x2, where A = 4.1×10−26 N·m2, and where
xis the distance from the alpha particle to the center of the nucleus. What is the speed of the alpha particle when it is (a) 4 nuclear radii from the nucleus; (b) 100 nuclear radii from the nucleus; (c) very far from the nucleus (x → ∞)?
Solution
In Problem 16 above, we found the potential energy difference for a repulsive inverse square force,
U (x2) − U(x1) = − Z x2 x1 A x2dx = A 1 x x2 x1 = A 1 x2 − 1 x1 ,
for x2< x1. If the kinetic energy of an alpha particle
at the nuclear surface is zero, K2= 0, then its kinetic
energy at some other point is K1= U2− U1.
Therefore, its speed is v1 =r 2K1 m = s 2A mx2 1 −xx2 1 = 2(4.1×10−26N·m2) (6.7×10−27 kg)(1.43×10−10 m) 1 −x2 x1 1/2 = (2.93×105 m/s)p1 − x2/x1,
where x2 is the nuclear radius. (a) If x1= 4x2, v1=
2.53×105 m/s, and (b) if x 1= 100x2, v1= 2.91× 105 m/s. (c) Evidently, 2.93×105m/s = v 1 for x1→ ∞.
Problem
76. A mass m is attached to a spring of constant k that is hanging from the ceiling. (a) Taking the zero of potential energy with the spring in its normal unstretched position, derive an expression for the total potential energy (gravitational plus spring) as a function of distance y taken as positive downward. (b) Find the point where the potential energy is a minimum, and explain its significance. (c) Find a second point where the total potential energy is zero. Discuss its significance in terms of an experiment where you attach the mass to the unstretched spring and let go.
Solution
(a) The elastic potential energy of the spring is 12ky2,
and the gravitational potential energy, relative to the unstretched height, is −mgy. The total potential energy is U (y) = 12ky2− mgy. (b) The minimum of
U (y) is found by setting its derivative equal to zero, dU/dy = ky − mg = 0, or y0= mg/k. Since Fy=
−dU/dy, y0 is the position of equilibrium. (c) The
zeros of U (y) are y = 0 and y = 2mg/k = 2y0. If the
mass is released from y = 0 at rest (K0= 0), the total
energy is E = 0, and the zeros of U (y) are the turning points of the motion.
Problem 76 (1) Solution.
Problem 76 (2) Solution.
Problem
77. With the brick of Problem 9 standing on end, what is the minimum energy that can be given the brick to make it fall over?
Solution
To fall over, the center of the brick must lie to the right of the vertical through its 8-cm edge, as shown. The potential energy of the brick would have to be increased by Ub− Ua= mg(yb− ya). If we assume that
energy is conserved between a and b, this is equal to the minimum kinetic energy sought (Ka− Kb= Ub −
Ua= Kamin, if Kb− 0). The numerical value is just the
difference between the answers to Problems 9(b) and 9(a), but recalculating, we find mg(yb− ya) =
(1.5 kg)(9.8 m/s2)1
2[p(20 cm)2+ (5.5 cm)2− 20 cm] =
Problem 77 Solution.
Problem
78. A bug lands on top of the frictionless, spherical head of a bald man. It begins to slide down the head (Fig. 8-47). Show that the bug leaves the head when it has dropped a vertical distance one-third the radius of the head.
figure 8-47 Problem 78.
Solution
The conservation of mechanical energy between points Oand P requires K0+ U0= K + U, or mgR = 12mv2+
mgy (since the bug starts from rest). Thus, v2= 2g×
(R − y) = 2gd. The bug leaves the surface when the normal force has decreased to zero. From the radial component of Newton’s second law, mg cos θ − N = mv2/R, so this occurs when N = mg cos θ − mv2/R =
0, or v2= gR cos θ = gy = g(R − d). Combining these
results, we find v2= 2gd = g(R − d), or d = 1 3R.
Problem
79. Together, the springs in a 1200-kg car have an effective spring constant of 110,000 N/m and can compress a maximum distance of 40 cm. What is
Problem 78 Solution.
the maximum abrupt drop in road level (Fig. 8-48) that the car can tolerate without “bottoming out”—that is, without its springs reaching
maximum compression? Assume the car is driving fast enough that it becomes temporarily airborne.
h
figure 8-48 Problem 79.
Solution
Let us neglect any possible losses in energy and assume that the kinetic energy associated with the car’s horizontal motion is unchanged by the drop. Then the energy principle implies that Ugrav(A) =
Ugrav(B) + Uspr(B), where A is a point before the drop
in the road, and B is the point after the drop where the springs are maximally compressed (the kinetic energy associated with the car’s vertical motion on its springs is instantaneously zero at B). Thus, Ugrav(A) −
Ugrav(B) = mgh = Uspr(B) = 12kx2, or h =
(110,000 N/m)(0.4 m)2/2(1200×9.8 N) = 74.8 cm.
Problem
80. Show that the rope in Example 8-6 will cease to be taut when the rope has caught on the rock and makes an angle θ = cos−1 2ℓ 3a cos θ0+a ℓ − 1 with the vertical. Show that your answer is consistent with that of Problem 33.
Solution
The condition for the rope to be taut going around the rock (at θ1 in Fig. 8-13) is 0 ≤ T = mg cos θ1+ mv12/a
(this follows from the radial component of Newton’s second law), or v2
1≥ −ga cos θ1. The conservation of
energy requires that K0+ U0= U0= K1+ U1, or
mgℓ(1 − cos θ0) =12mv12+ mga(1 − cos θ1) (as in
Example 8-6). When these results are combined, v12= 2gℓ(1 − cos θ0) − 2ga(1 − cos θ1) ≥ −ga cos θ1, or
cos θ1≥ (2ℓ/3a)(cos θ0− 1 + a/ℓ), which is the limiting
value stated. If θ0= 90◦and a = 25ℓ, then θ1= cos−1×
[2 3· 5 2(0 − 1 + 2 5)] = cos
−1(−1) = 180◦; i.e., the rope is
taut at the top of the small circle.
Problem
81. An electron with kinetic energy 0.85 fJ enters a region where its potential energy as a function of position is U = ax2− bx, where a = 2.7 fJ/cm2
and b = 4.2 fJ/cm. (a) How far into the region does the electron penetrate? (b) At what position does the electron have its maximum speed? (c) what is this maximum speed?
Solution
(a) The electron’s total energy, E = 0.85 fJ, equals its initial kinetic energy upon entering the region at x = 0. At any other point in the region x ≥ 0, E = K + U, so the point of maximum penetration (the turning point) can be found from the equation K(xm) = E −
U (xm) = 0. Therefore, ax2m− bxm− 0.85 fJ = 0, or
xm= [4.2 +p(4.2)2+ 4(2.7)(0.85)]/2(2.7) = 1.74 cm.
(The negative root can be discarded since x ≥ 0.) (b) The maximum kinetic energy occurs at the point where U is a minimum; that is, dU/dx = 2ax0− b = 0,
or x0= b/2a = 0.778 cm. (c) vmax=p2Kmax/me =
p2(E − Umin)/me. Now, Umin= x0(ax0− b) =
−b2/4a = −(4.2 fJ/cm)2/4(2.7 fJ/cm2 ) = −1.63 fJ, so that vmax=p2(0.85 + 1.63) fJ/(9.11×10−31kg) = 7.38×107 m/s. Problem 81 Solution.
Problem
82. A particle of mass m is subject to a force F = (a√x)ˆı, where a is a constant. The particle is initially at rest at the origin, and is given a slight nudge in the positive x direction. Find an
expression for the particle’s speed as a function of position x.
Solution
The force is conservative, so K0+ U0= K + U, where
the subscript 0 refers to the origin, and K0= 0. The
potential energy difference is U − U0= − ∫0xa
√ x′×
dx′= −23ax 3/2
, so the speed for points x ≥ 0, from the kinetic energy, is v =p2K/m = p2(U0− U)/m =
2(a/3m)1/2x3/4.
Problem
83. (a) Repeat the previous problem for the case of a force F = (ax − bx3)ˆı, where a and b are positive
constants. (b) What is the significance of the negative square root that can occur for some values of x? (c) Find an expression for the particle’s maximum speed.
Solution
(a) With the same reasoning in the previous problem’s solution (i.e., conservation of energy K0+ U0= K + U,
with K0= 0 at x = 0), U − U0= − ∫0x(ax′− bx′3)dx′ =
−(x2/2)(a − bx2/2) = −K = −1 2mv
2, or v =
p(x2/m)(a − bx2/2). (b) The total energy of the
particle is E = U0, so the particle is confined to the
region where U (x) ≤ E. (This is the same as the condition K ≥ 0.) U(x) = E when x = ±p2a/b, which are the turning points of the motion. Values of position with |x| >p2a/b are forbidden, classically. (c) The maximum speed occurs at the minimum of potential energy. dU/dx = bx3− ax = 0 has roots at x = 0 and x = ±pa/b. The former is a maximum and the latter are minima of U. Substitution into the expression for the speed yields vmax=
p(a/bm)(a − ba/2b) = a/√2bm.
Problem
84. A 17-m-long vine hangs vertically from a tree on one side of a 10-m-wide gorge, as shown in Fig. 8-49. Tarzan runs up, hoping to grab the vine, swing over the gorge, and drop vertically off the vine to land on the other side. At what minimum speed must he be running?
Solution
If we consider Tarzan as a simple pendulum, we may apply the reasoning of Example 8-6(a) to find his
