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AP Physics B: Work/Energy/Power Homework Challenge Sheet

1. A 285-kg load is lifted 22.0 m vertically with an acceleration a=0.160g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

2. A cyclist intends to cycle up a 7.8º hill whose vertical height is 150 m. Assuming the mass of bicycle plus cyclist is 75 kg, (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves the bike 5.1 m along its path, calculate the average force that must be exerted on the pedals tangent to their circular

3. An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

4. A 0.620-kg wood block is firmly attached to a very light horizontal spring

(

k=180N m

)

as shown. It is noted

that the block–spring system, when compressed 5.0 cm and released, stretches out 2.3 cm beyond the equilibrium position before stopping and turning back. What is the coefficient of

kinetic friction between the block and the table?

5. A 280-g wood block is firmly attached to a very light horizontal spring. The block can slide along a table where the coefficient of friction is 0.30. A force of 22 N

compresses the spring 18 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

6. Early test flights for the space shuttle used a “glider” (mass of 980 kg including pilot) that was launched horizontally at 500km h from a height of 3500 m. The glider eventually landed at a speed of 200km h. (a)

What would its landing speed have been in the absence of air resistance? (b) What was the average force of air resistance exerted on it if it came in at a constant glide of 10º to the Earth?

7. What minimum horsepower must a motor have to be able to drag a 310-kg box along a level floor at a speed of 1.20m s if the coefficient of friction is 0.45?

8. A bicyclist coasts down a 7.0º hill at a steady speed of 5.0ms. Assuming a total mass of 75 kg (bicycle plus

(2)

Solutions:

1. (a) From the free-body diagram for the load being lifted, write Newton’s 2nd law for

the vertical direction, with up being positive.

(

)

(

)

T

2 3

T

0.160

1.16 1.16 285 kg 9.80 m s 3.24 10 N

F F mg ma mg

F mg

= − = = →

= = = ×

(b) The net work done on the load is found from the net force.

(

)

(

)

(

)

(

)

o 2

net net

3

cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m

9.83 10 J

W =F d = mg d =

= ×

(c) The work done by the cable on the load is

(

)

(

)

(

)

(

)

o 2 4

cable T cos 0 1.160 1.16 285 kg 9.80 m s 22.0 m 7.13 10 J

W =F d = mg d = = ×

(d) The work done by gravity on the load is

(

)

(

)

(

)

o 2 4

G cos180 285 kg 9.80 m s 22.0 m 6.14 10 J

W =mgd = −mgd= − = − ×

(e) Use the work-energy theory to find the final speed, with an initial speed of 0.

(

)

2 2

1 1

net 2 1 2 2 2 1

3

2

2 1

2 9.83 10 J 2

0 8.31m s

285 kg

net

W KE KE mv mv

W

v v

m

= − = − →

×

= + = + =

2. (a) The work done against gravity is the change in PE.

(

) (

)

(

2

)

(

)

5

against 2 1

gravity

75 kg 9.8 m s 150 m 1.1 10 J

W = ∆PE=mg yy = = ×

(b) The work done by the force on the pedals in one revolution is equal to the tangential force times the circumference of the circular path of the pedals. That work is also equal to the energy change of the bicycle during that revolution. Note that a vertical rise on the incline is related to the distance along the incline by rise=distance sin

(

θ

)

.

( )

(

)

(

)

(

)

(

)

pedal tan 1 rev 1 rev 1 rev

force

2 o

2 1 rev

tan

2 sin

75 kg 9.8 m s 5.1 m sin 7.8 sin

4.5 10 N

2 2 0.18 m

W F r PE mg y mgd

mgd F

r

π θ

θ

π π

= = ∆ = ∆ = →

= = = ×

3. The maximum acceleration of 5.0 g occurs where the force is at a maximum. The maximum force occurs at the bottom of the motion, where the spring is at its maximum compression. Write Newton’s 2nd law for the elevator at the bottom of the motion, with up

as the positive direction.

net spring 5.0 spring 6.0

F =FMg=Ma= MgF = Mg

T

F

mg

mg

spring

(3)

Now consider the diagram for the elevator at various points in its motion. If there are no non-conservative forces, then mechanical energy is conserved. Subscript 1 represents the elevator at the start of its fall, and subscript 2 represents the elevator at the bottom of its fall. The bottom of the fall is the zero location for gravitational PE

(

y=0

)

. There is also a point at the

top of the spring that we will define as the zero location for elastic PE (x = 0). We have v1 =0,

1

y = +x h, x1=0, v2 =0, y2 =0, and x2 =x. Apply conservation of energy.

(

)

(

)

2 2 2 2

1 1 1 1

1 2 2 1 1 2 1 2 2 2 2 2

2 2

1 1

2 2

2 2

2

1 2

0 0 0 0

6.0 6 6 12

6.0

spring

E E Mv Mgy kx Mv Mgy kx

Mg x h kx Mg x h kx

Mg Mg Mg Mg

F Mg kx x Mg h k k

k k k h

= → + + = + + →

+ + + = + + → + =

= = → = →

+

=

→ =

4. Use conservation of energy, including the non-conservative frictional force. The block is on a level surface, so there is no gravitational PE change to consider. The frictional force is given by

fr k N k

FFmg, since the normal force is equal to the weight. Subscript 1 represents the

block at the compressed location, and subscript 2 represents the block at the maximum stretched position. The location of the block when the spring is neither stretched nor

compressed is the zero location for elastic PE (x = 0). Take right to be the positive direction. We have v1 =0, x1= −0.050 m, v2 =0, and x2 =0.023 m.

(

)

(

) (

) (

)

(

)

(

)

(

)

o 1 2 1 2 1 2 1 2

NC 1 2 fr 2 1 2 1 2 2 2 2

2 2

1 1

1 2

2 2

2 2

2 2

1 2

2

cos180

180 N m 0.050m 0.023m

0.40

2 2 0.620 kg 9.80 m s 0.073 m

k

k

W E E F x mv kx mv kx

mg x kx kx

k x x

mg x µ

µ

+ = → ∆ + + = + →

− ∆ + = →

− −

= = =

 

 

5. Use conservation of energy, including the non-conservative frictional force. The block is on a level surface, so there is no gravitational PE change to consider. Since the normal force is equal to the weight, the frictional force is FfrkFNkmg. Subscript 1 represents the block at the

compressed location, and subscript 2 represents the block at the maximum stretched position. The location of the block when the spring is neither stretched nor compressed is the zero location for elastic PE (x = 0). Take right to be the positive direction. We have v1 =0,

1 0.18 m

x = − , and v2 =0. The value of the spring constant is found from the fact that a 22-N Start of fall

Contact with spring,

Bottom of fall, 0 for gravitational PE h

(4)

force compresses the spring 18 cm, and so k =F x=22 N 0.18 m=122.2 N m. The value of

2

x must be positive.

(

)

(

)(

)(

)

(

)(

)(

) (

) (

)

o 1 2 1 2 1 2 1 2

NC 1 2 fr 2 1 2 1 2 2 2 2

2 2 2 2

1 1

2 1 2 1 2 2 2 2 1 1

2 2

2 2

2 2

cos180

2 2

0

2 0.30 0.28 9.80 2 0.30 0.28 9.80

0.18 0.18 0

122.2 122.2

0.01347

k k

k

W E E F x mv kx mv kx

mg mg

mg x x kx kx x x x x

k k

x x

x x

µ µ

µ

+ = → ∆ + + = + →

− − + = → + − + = →

+ − − + − = →

+

2−0.02997=0 → x2 =0.1665 m,−0.1800m → x2 = 0.17 m

6. (a) If there is no air resistance, then conservation of mechanical energy can be used. Subscript 1 represents the glider when at launch, at subscript 2 represents the glider at landing. The landing location is the zero location for elastic PE (x = 0). We have y1 =500 m, y2 =0, and

1

1m s

500 km h 138.9 m s

3.6 km h

v =

=

. Solve for v2

(

)

(

)

(

)

2 2

1 1

1 2 2 1 1 2 2 2

2

2 2

2 1 1

3

3.6 km h

2 138.9 m s 2 9.80 m s 3500 m 296 m s

1m s

1067 km h 1.1 10 km h

E E mv mgy mv mgy

v v gy

= → + = + →

= + = + =

= ≈ ×

(b) Now include the work done by the non-conservative frictional force. Consider the diagram of the glider. Calculate the work done by friction.

o

NC fr fr fr o

3500 m cos180

sin10

W =F d = −F d= −F

Use the same subscript representations as above, with y1 , v1, and y2 as before, and

2

1m s

200 km h 55.56 m s

3.6 km h

v =

=

. Write the energy conservation equation and solve for

the frictional force.

7. Draw a free-body diagram for the box being dragged along the floor. The box has a constant speed, so the acceleration is 0 in all directions. Write Newton’s 2nd law for both the x

(horizontal) and y (vertical) directions.

N N

P fr P fr N

0

0

y

x k k

F F mg F mg

F F F F F µ F µ mg

= − = → =

= − = → = = =

The work done by FP in moving the crate a distance ∆x is given by W = ∆FP xcos 0o =µkmg x∆ .

(5)

(

)(

)

(

2

)

(

)

0.45 310 kg 9.80 m s 1.20 m s 1641W

1 hp

1641W 2.2 hp

746 W

k

k k x

mg x

W x

P mg mgv

t t t

µ

µ µ

∆ ∆

= = = = = =

=

 

 

 

8. First, consider a free-body diagram for the cyclist going down hill. Write Newton’s 2nd law for

the x direction, with an acceleration of 0 since the cyclist has a constant speed.

fr fr

sin 0 sin

x

F =mg θ−F = → F =mg θ

Now consider the diagram for the cyclist going up the hill. Again, write Newton’s 2nd law for the

x direction, with an acceleration of 0.

fr P sin 0 P fr sin

x

F =FF +mg θ = → F =F +mg θ

Assume that the friction force is the same when the speed is the same, so the friction force when going uphill is the same magnitude as when going downhill.

P fr sin 2 sin

F =F +mg θ = mg θ

The power output due to this force is given by Eq. 6-17.

(

)

(

2

)

(

)

o

P

2

2 sin 2 75 kg 9.8 m s 5.0 m s sin 7.0

9.0 10 W

P=F v= mgv θ =

References

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