Unit 2 Energy and Momentum
Unit 2 Energy and Momentum
ARE YOU READY?
ARE YOU READY?
(Pages 174–175) (Pages 174–175)
Knowledge and Understanding
Knowledge and Understanding
1.1. Some forms Some forms of energy of energy are the are the chemical potential chemical potential energy stored energy stored in the in the gasoline, kinetic gasoline, kinetic energy, gravitational energy, gravitational potentialpotential energy, thermal energy (heat from the motor), and sound.
energy, thermal energy (heat from the motor), and sound. 2. (a) mass 2. (a) mass (b) energy (b) energy (c) force (c) force (d) power (d) power 3.
3. joule joule – – James James JouleJoule newton – Isaac Newton newton – Isaac Newton watt – James Watt watt – James Watt 4.
4. The work The work you you do on do on the box the box depends on depends on its mass,its mass, mm, the gravitational field strength,, the gravitational field strength, g g , and the vertical height it is raised,, and the vertical height it is raised,
∆ ∆ y y..
Inquiry and Communication
Inquiry and Communication
5. (a) 5. (a) 2299..55 2255..11 101000%% 1144..99%% 29.5 29.5 − − × × ==The percentage of initial kinetic energy lost is 14.9%. The percentage of initial kinetic energy lost is 14.9%. (b)
(b) The energy did not cease to exist, but was The energy did not cease to exist, but was converted into other forms (likely heat and sound).converted into other forms (likely heat and sound). 6.
6. (a) (a) The measure the The measure the efficiency of this incline, you would efficiency of this incline, you would need to know need to know the distance up the the distance up the ramp, the force applied, theramp, the force applied, the mass of the sled and child and the vertical height. If you knew the angle that the hill was compared to the horizontal, mass of the sled and child and the vertical height. If you knew the angle that the hill was compared to the horizontal, you could calculate the vertical height from the length up the slope using trigonometry.
you could calculate the vertical height from the length up the slope using trigonometry. (b)
(b) For 100% efficiency, there could be no friction between the sled and thFor 100% efficiency, there could be no friction between the sled and the ground.e ground. 7.
7. The conservation The conservation of energy refers of energy refers to the inability to to the inability to create or destroy create or destroy energy during an energy during an interaction. Conserving energy interaction. Conserving energy isis the reduction of our use of energy converted from fossil fuels and other sources to allow them to last longer.
the reduction of our use of energy converted from fossil fuels and other sources to allow them to last longer.
Making Connections
Making Connections
8.8. (a) (a) The large mass The large mass of the bus help to of the bus help to reduce injury becausreduce injury because the acceleration will be small. The e the acceleration will be small. The bumpers and body bumpers and body panelspanels also help to absorb energy to minimize the energy that will be absorbed by passengers.
also help to absorb energy to minimize the energy that will be absorbed by passengers. (b)
(b) We could increase the safety of buses We could increase the safety of buses by adding more padding around passenby adding more padding around passenger compartments, seat belts, and perhapsger compartments, seat belts, and perhaps even air bags.
even air bags. (c)
(c) School busses currently have exceSchool busses currently have excellent safety ratings, and the added features in part (b) are expensive to install.llent safety ratings, and the added features in part (b) are expensive to install. 9.
9. (a) (a) The ski jumper will acquThe ski jumper will acquire gravitational potential energy as he ire gravitational potential energy as he is transported to the top of is transported to the top of the hill. This gravitationalthe hill. This gravitational energy will be converted to kinetic energy as the jumper slides down the hill, and then reconverted back into some energy will be converted to kinetic energy as the jumper slides down the hill, and then reconverted back into some gravitational energy as he jumps from the end of the ramp. All of the energy will be converted to kinetic as he reaches gravitational energy as he jumps from the end of the ramp. All of the energy will be converted to kinetic as he reaches the bottom of the hill (if we ignore the loss due to friction).
the bottom of the hill (if we ignore the loss due to friction). (b)
(b) The law of consThe law of conservation of energy is whervation of energy is what governs the energy at governs the energy transformations.transformations. (c)
(c) The main safety feature of this sport is the careful construction of the hill so that the ski jumper does The main safety feature of this sport is the careful construction of the hill so that the ski jumper does not experience anot experience a large normal force from landing. Also the protection of the head by a helmet and suits minimize injury in case of a fall. large normal force from landing. Also the protection of the head by a helmet and suits minimize injury in case of a fall.
Math Skills
Math Skills
10. (a) 10. (a) W W ∝ ∝ ∆∆d d (b) (b) E E PP ∝∝xx22 (c) (c) P P 11 t t ∝ ∝ (d) (d) E E K K ∝∝vv22The graphs for (a)–(d) are shown below.
11. (a) x-components: F K , F Acos α y-components: F N , F Asinα , F g (b) x-components: F K , F Acos α , F gsinβ
y-components: F N, F Asinα , F gcosβ 12.
Technical Skills and Safety
13. (a) Place the air pucks on the table while it is running to see if they regularly move in one direction.
(b) The determination of speeds depends on the assumption that the speeds are constant. For a table that is not level, a component of the force of gravity will be accelerating the objects and changing their speed and direction.
(c) Other sources of error can be vibration from motion, and air currents from a location near a ventilation duct for the school environmental controls. You must also check equipment to make sure there are no rough edges that might catch. 14. Most sensors are electronic and have a shock hazard so the electrical equipment should be kept away from water. This is
also wise to protect the equipment. All experiments should be run slowly in a trial form to make sure that any electrical connectors or cables do not interfere with data collection or damage the electronic devices.
CHAPTER 4 WORK AND ENERGY
Reflect on Your Learning
(Page 176)1. A compression spring is designed to be at its maximum length when there is no force, and resist being squeezed together. An extension spring is designed to be at its minimum length when there is no force, and resist being stretched. Some examples of compression springs are the springs in a car and a retractable ballpoint pen. Some examples of extension springs are bungee cords and storm door springs.
2. During the bounce of the ball, most of the kinetic energy is stored as elastic potential energy to be converted back into kinetic energy after the bounce. Some energy is “lost” (converted) to unwanted forms such as heat and sound.
3. A grandfather clock uses the gravitational potential energy to operate a pendulum with a constant period. 4. (a) The work done on each cart is the same.
(b) The work done on each box would be the same since it only depends on the size of the component of force in the direction of motion and the displacement. For each box, these values are still the same.
5.
6. The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slow air release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involved would mean the choice of a strong material, such as steel, for strength.
Try This Activity: Which Ball Wins?
(Page 177)(a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speed up more quickly than the ball on track X.
(b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another.
4.1 WORK DONE BY A CONSTANT FORCE
PRACTICE
(Pages 181–182)Understanding Concepts
1. F 1 will do more work than F 2 because the component of F 2 in the direction of motion is smaller than F 1.
2. No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is always opposite the direction of motion, the kinetic friction will always do negative work.
3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object. 4. m = 2.75 kg (a) ∆d = 1.37 m W = ? ( cos ) ( cos ) (2.75 kg)(9.80 N/kg)(cos 0 )(1.37 m) 36.9 J W F d mg d W θ θ = ∆ = ∆ = ° =
The work done to move the plant 1.37 m up is 36.9 J. (b) ∆ y = 1.07 m
µK = 0.549 W = ?
First we must calculate the normal force acting on the potted plant:
N g N g 0 0 (2.75 kg)(9.80 N/kg) y y F ma F F F F mg Σ = = − = = = =
Let F A be the applied force to move the potted plant horizontally: A K A K N A 0 0 (0.549)(26.95 N ) 14.796 N x x F ma F F F F F F µ
Σ =
=
−
=
=
=
=
=
To calculate the work done on the potted plant: A ( cos ) (14.796 N)(cos0 )(1.07 m) 15.8 J W F d W θ
=
∆
=
°
=
The work done to move the plant 1.07 m across the shelf is 15.8 J. 5. m = 24.5 kg
F
= 14.2 N [22.5
°
below the horizontal]∆d = 14.8 m W = ?
We only need to consider the component of force in the direction of motion: ( cos ) (14.2 N)(cos22.5 )(14.8 m) 194 J W F d W θ
=
∆
=
°
=
The work done by the force is 194 J. 6. F T
=
12.5 N [19.5 above the horizontal]°
W = 225 J ∆d = ? ( cos ) cos 225 J (12.5 N)(cos19.5 ) 19.1 m W F d W d F d θ θ
=
∆
∆ =
=
°
∆ =
The toboggan moves 19.1 m.
7. (a) We will calculate the area using the formula for a rectangle
(4.0 N)(2.0 m) 8.0 J A lw A
=
=
=
The area represents the work done on the object.
(b) First calculate the area of the second portion of the graph:
( 4.0 N)(6 0 m 2.0 m) 8.0 J A lw . A
=
= −
−
= −
The total work is 8.0 + (–8.0) = 0.0 J
(c) One situation could be pushing a box across a table and pulling it back.
Applying Inquiry Skills
8. You could set the pen on the paper and pull the paper across the desk. The force of static friction between the paper and the pen is in the direction of motion, doing positive work on the pen.
Understanding Concepts
9. Four different situations are:•
a book sitting on a desk (∆
d = 0)•
a student carrying a book at a constant speed (θ = 90º)•
a teacher whirling a putty pat in a circle at the end of a string•
a toy car travelling in a circular path10. (a) A box being pulled by string at an angle involves the forward component doing positive work, and the vertical component doing zero work.
(b) A box being pulled up a ramp involves the parallel component of gravity doing negative work, and the perpendicular component of gravity doing zero work.
Section 4.1 Questions
(Page 183)Understanding Concepts
1. The everyday use of the word “work” is different from the usage in physics when it is referring to employment, or a duty to perform. “Working” as a teacher involves very little physical work. The physics definition of work means the transfer of energy to an object to move it a certain distance. Many types of employment or daily activities involve physical work. For example, the sentence “Loading the cement bags onto the truck was a lot of work,” uses the word “work” similar to the physics definition of work.
2. The centripetal force is always directed toward the centre of the circle, and is by definition perpendicular to the motion of the object. The 90º angle means that work is not done on the object by the centripetal force.
3. As you push on a wall, you are exerting a force, which involves the use of energy. Even though no physical work is being done, your muscles are still burning your body’s fuel, causing you to become tired.
4. Assuming the classroom to be 4 m tall, and the student to have a mass of 70 kg:
(
)
3 ( cos ) cos (70 N)(9.80 N/kg)(cos 0 )(4.0 m) 3.0 10 J W F d mg d W θ θ=
∆
=
∆
=
°
=
×
It would take about 3.0 103 J, or 3.0 kJ of work to climb the ladder. 5. F A
(a) The work done by the boy (W B): B T B ( cos ) (75 N)(cos 32 )(13 m) 826.8 J W F d W θ
=
∆
=
°
=
The work done by the girl (W G): G A G ( cos ) (75 N)(cos 22 )(13 m) 904.0 J W F d W θ
=
∆
=
°
=
The total work done:
total B G 3 total 828.8 J 904.0 J 1.73 10 J W W W W
=
+
=
+
=
×
The total amount of work done on the crate is 1.7 10 –3 J.
(b) The crate is moving at a constant speed, so it is not gaining any energy. This means that the crate will have the same amount of energy before and after the move, so it must have work done on it opposite the direction of motion. Therefore, the work done on the crate by the floor is –1.7 103 J.
6. W = 9.65 102 J
∆d = 45.3 m F
= 24.1 N [parallel to the handle of the sleigh]
θ = ? 1 1 ( cos ) cos cos 965 J cos (24.1 N)(45.3 m) 27.9 W F d W F d W F d θ θ θ θ − −
=
∆
=
∆
=
∆
=
=
°
The angle between the snowy surface and the handle is 27.9º. 7. (a) ∆d = 38 m
m = 66 kg A
F
= 58 N [18
°
above the horizontal] F N = ?µK = ?
First we must calculate the normal force:
N A g N g A A N 0 sin18 0 sin18 sin18 (66 kg)(9.80 N/kg) (58 N)sin18 628.88 N y y F ma F F F F F F mg F F
Σ =
=
+
° − =
= −
°
=
−
°
=
−
°
=
We then calculate the force of gravity: A K K A K 0 cos18 0 cos18 (58 N) cos18 55.161 N x x F ma F F F F F
Σ =
=
° −
=
=
°
=
°
=
The coefficient of kinetic friction is: K K N K 55.161 N 628.88 N 0.088 F F µ µ
=
=
=
The normal force is 6.3 102 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088.
(b) W = ? 3 ( cos ) (55.161 N)(cos180 )(38 m) 2.1 10 J W F d W θ
=
∆
=
°
= − ×
The work done by kinetic friction is –2.1 103 J.
(c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan. (d) ∆d = 25 m W = ? 3 ( cos ) (58 N)(cos18 )(25 m) 1.4 10 J W F d W θ
=
∆
=
°
= ×
The work done by the parent is 1.4 103 J.
Applying Inquiry Skills
8. As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for angles equal to 90º, and negative for angles between 90º and 180º.
Making Connections
9. Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermal energy.
4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM
PRACTICE
(Pages 186–187)2. The kinetic energy is proportional to the speed, so if the speed increases by (a) 2, the kinetic energy will increase by a factor of 22, or 4
(b) 3, the kinetic energy will increase by a factor of 32, or 9
(c) 37%, the kinetic energy will increase by a factor of 1.372, or 1.9 3. Assume a 75-kg student running at 8.0 m/s:
2 K 2 3 K 1 2 1 (75 kg)(8.0 m/s) 2 2.4 10 J E mv E
=
=
=
×
The kinetic energy at maximum speed is 2.4 kJ. 4. m = 45 g = 4.5 10 –2 kg vi = 0 m/s vf = 43 m/s (a) W = ?
(
) ( )
(
)
K 2 2 f i 2 2 2 1 ( ) 2 1 (4.5 10 kg) 43 m/s 0 m/s 2 41.6025 J, or42 J W E m v v W −= ∆
=
−
=
×
−
=
The work done by the club is 42 J. (b) ∆d = 2.0 cm = 2.0 10 –2 m F = ? 2 3 41.6025 J 2.0 10 m 2.1 10 N W F d W F d F −
= ∆
=
∆
=
×
=
×
The average force exerted by the club is 2.1 × 103 N. 5. m = 27 g = 2.7 10 –1 kg F = 95 N ∆d = 31 cm = 3.1 10 –1 m vf = ? K 2 f 2 f f 1 1 f 1
(since the initial speed is zero) 2 2 2 2(95 N)(3.1 10 m) 2.7 10 kg 47 m/s W E F d mv F d v m F d v m v − −
= ∆
∆ =
∆
=
∆
=
×
=
×
=
The final speed of the arrow is 47 m/s. 6. m = 4.55 104 kg
vi = 1.22 104 m/s F = 3.85 105 N
∆d = 2.45 106 m vf = ?
K 2 2 f i 2 2 f i 2 f i 5 6 4 2 4 4 f 1 1 2 2 1 1 2 2 2 2(3.85 10 N)(2.45 10 m) (1.22 10 m/s) 4.55 10 kg 1.38 10 m/s W E F d mv mv mv F d mv F d v v m v
= ∆
∆ =
−
= ∆ +
∆
=
+
×
×
=
+
×
×
=
×
The final speed of the probe is 1.38 104 m/s. 7. m = 28.0 kg
A F
= 95.6 N [35
°
above the horizontal] F K = 75.5 Nvi = 0 m/s
∆d = 0.750 m vf = ?
The total work done on the box will become kinetic energy. Since the initial speed is zero:
(
)
(
)
(
)
2 f 2 A K f 2 f A K f A K f 1 2 1 cos 35.0 cos180 2 2 cos 35.0 cos180 2 cos 35.0 cos180 2(0.750 m) (95.6 N)(0.81952) (75.5 N)( 1) 20.8 kg 0.45 m/s W mv F d F d mv d v F F m d v F F m v=
°∆ +
°∆ =
∆
=
° +
°
∆
=
° +
°
=
+
−
=
The final speed of the box is 0.45 m/s. 8. W = 1.47(cos 38º ) = 1.16
The toboggan would have increased its kinetic energy by 16%.
Applying Inquiry Skills
9. W
= ∆
F d 2 2 2 N m kg m m s kg m s W= ⋅
⋅
=
⋅
⋅
=
2 K 2 2 K 2 1 2 m kg s kg m s E mv E=
= ⋅
⋅
=
The base units for both are the same.
Making Connections
10. m = 6.85 103 kg vA = 2.81 103 m/s vB = 8.38 103 m/s W = ?
(a) The work done is equal to the change in kinetic energy
(
)
K 2 1 2 B 2 A 2 2 B A 3 3 2 3 2 11 1 2 1 ( ) 2 1 (6.85 10 kg) (8.38 10 m/s) (2.81 10 m/s) 2 2.13 10 J W E mv mv m v v W= ∆
=
−
=
−
=
×
×
−
×
=
×
(b) The work done by Earth to move the satellite from A to B is 2.13 × 1011 J.
(
)
K 2 2 A B 2 2 A B 3 3 2 3 2 11 1 1 2 2 1 ( ) 2 1 (6.85 10 kg) (2.81 10 m/s) (8.38 10 m/s) 2 2.13 10 J W E mv mv m v v W= ∆
=
−
=
−
=
×
×
−
×
= −
×
The work done by Earth to move the satellite from B to A is –2.13 × 1011 J.
Section 4.2 Questions
(Page 188)
Understanding Concepts
1. The first doubling will require much less energy than the second doubling of the speed. This can clearly be shown using: 2 2 f i 2 2 f i 1 ( ) 2 W m v v W v v
=
−
∝ −
To go from v to 2v: 2 2 f i 2 2 2 2 2 (2 ) ( ) 4 3 W v v v v v v W v= −
=
−
=
−
=
To go from 2v to 4v: 2 2 f i 2 2 2 2 2 (4 ) (2 ) 16 4 11 W v v v v v v W v= −
=
−
=
−
=
The first doubling of speed will require work proportional to 3 times the square of the original speed. The second doubling will require work proportional to 11 times the square of the original speed.
2. m = 1.50 103 kg v = 18.0 m/s [E] E K = ? 2 K 3 2 5 K 1 2 1 (1.5 10 kg)(18.0 m/s) 2 2.43 10 J E mv E
=
=
×
=
×
The kinetic energy of the car is 2.43 105 J.
3. (a) v = 1.150 × 18.0 = 20.7 m/s E K = ? 2 K 3 2 5 K 1 2 1 (1.5 10 kg)(20.7 m/s) 2 3.21 10 J E mv E
=
=
×
=
×
The new kinetic energy of the car is 3.21 105 J. (b) The increase in E K is:
5 5 3.21 10 J 1.32 2.43 10 J
×
=
×
We can verify this with (1.15)2 = 1.32
This represents an increase in the kinetic energy of 32%. (c) W = ? K 5 5 4 3.21 10 J 2.43 10 J 7.8 10 J W E W
= ∆
= × − ×
=
×
The work done to speed up the car was 7.8 104 J.
4. m = 55 kg E K = 3.3 103 J v = ? 2 K K 3 1 2 2 2(3.3 10 J) 55 kg 11 m/s E mv E v m v
=
=
×
=
=
The speed of the sprinter is 11 m/s. 5. v = 12 m/s E K = 43 J m = ? 2 K K 2 2 1 2 2 2(43 J) (12 m/s) 0.60 kg E mv E m v m
=
=
=
=
6. m = 0.353 kg ∆d = 89.3 cm = 0.893 m (a) W = ? ( cos ) (0.353 kg)(9.80 N/kg)(cos 0 )(0.893 m) 3.0892 J W mg d W θ
=
∆
=
°
=
The work done by gravity is 3.09 J.
(b) Using the work energy theorem,W
= ∆
E K . Since vi = 0: 2 f f f 1 2 2 2(3.0892 J ) 0.353 kg 4.18 m/s W mv W v m v=
=
=
=
The speed of the plate just before it hits the floor is 4.18 m/s. 7. m = 61 kg θ = 23
°
F K = 72 N vi = 3.5 m/s ∆d = 62 m vf = ?The component of gravity along the slope is mg sin 23º. Using the work energy theorem:
2 2 K f i 2 2 f K i 2 K 2 f i 2 K f i 2 2 f 1 1
sin 23 (cos 0 ) (cos180 )
2 2 1 1 sin 23 (1) ( 1) 2 2 2 2 sin 23 2 2 sin 23 2(72 N)(62 m) 2(9.8 m/s )(sin 23 )(62 m) (3.5 m/s) 61 kg 18 m/s mg d F d mv mv mv mg d F d mv F d v g d v m F d v g d v m v
°
°
∆ +
° ∆ =
−
=
° ∆ + − ∆ +
∆
=
°∆ −
+
∆
=
°∆ −
+
=
°
−
+
=
The speed of the skier after travelling 62 m downhill is 18 m/s. 8. m = 55.2 kg
∆d = 4.18 m µK = 0.27
Using the FBD to calculate F N, N N N 0 0 (55.2 kg)(9.80 N/kg) 540.96 N y y F ma F mg F mg F
Σ =
=
−
=
=
=
=
To calculate F K : K K N K (0.27)(540.96 N ) 146.06 N F F F µ=
=
=
Using the work-energy theorem:
K 2 2 K f i 1 (cos180 ) ( ) 2 W E F d m v v
= ∆
° ∆ =
−
Since vf = 0, 2 K i K i i 1 (cos180 ) 2 2 (cos180 ) 2(146.06 N)( 1)(4.18 m) 55.2 kg 4.7 m/s F d mv F d v m v° ∆ = −
−
° ∆
=
−
−
=
=
The initial speed of the skater was 4.7 m/s.
Applying Inquiry Skills
9. (a)
Car Speed (m/s) Car Energy (J) Truck Speed (m/s) Truck Energy (J)
10.0 6.0 × 104 10.0 2.5 × 107
20.0 2.4 × 105 20.0 1.0 × 108
30.0 5.4 × 105 30.0 2.2 × 108
40.0 9.6 × 105 40.0
4.0 × 108
To calculate the energy of the car and the truck, use the equation K 1 2 2 E
=
mv . To convert tonnes to kilograms, multiply by 1000:m = 1.2 t = 1.2 103 kg (for the car)
m = 5.0 102 t = 5.0 105 kg (for the truck) (b)
(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kinetic energy increases proportional to the square of the speed.
Making Connections
10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat.
(b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts. The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit, causing death.
4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACE
PRACTICE
(Page 191)Understanding Concepts
1. The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the way up, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you could argue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.
2. m = 62.5 kg
∆ y = 346 m (using the ground as y = 0) E g = ? g 5 g (62.5 kg)(9.80 N/kg)(346 m) 2.12 10 J E mg y E
∆ =
∆
=
∆ =
×
Relative to the ground, the gravitational potential energy is 2.12 105 J. 3. m = 58.2 g = 5.82 10 –2 kg
∆ y = 1.55 m (a) ∆ E g = ?
At 1.55 m above the court: g 2 g (5.82 10 m)(9.80 N/kg)(1.55 m) 0.884 J E mg y E −
∆ =
∆
=
×
∆ =
At the court height: g 2 g (5.82 10 m)(9.80 N/kg)(0.00 m) 0.00 J E mg y E −
∆ =
∆
=
×
∆ =
The gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is 0.00 J. (b) W = ? 2 ( cos ) ( cos ) (5.82 10 kg)(9.80 N/kg)(cos0 )(1.55 m) 0.884 J W F d mg d W θ θ −
=
∆
=
∆
=
×
°
=
At the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball. (c) The work done in (b) is equal to the change in kinetic energy of the ball.
4. m = 68.5 kg
m = 2.56 km = 2.56 103 m
θ = 13.9
°
First, calculate the vertical lift: 3 3 sin13.9 2.56 10 m 2.56 10 m(sin 13.9 ) 614.98 m y y y −
∆
° =
×
∆
=
×
°
∆ =
Then calculate the gravitational potential energy: g 5 g (68.5 kg)(9.80 N/kg)(614.98 m) 4.13 10 J E mg y E
∆ =
∆
=
∆ =
×
The skier’s gravitational potential energy at the top of the mountain is 4.13 105 J. 5. ∆ y = 2.36 m
∆ E g = –1.65 10 3
J m = ?
The ground is 2.36 m down from the pole, therefore: g g 3 1.65 10 J (9.80 N/kg)( 2.36 m) 71.3 kg E mg y E m g y m
∆ =
∆
∆
=
∆
−
×
=
−
=
The mass of the jumper is 71.3 kg.
6. (a) The first coin does not need to be lifted, so no work is done on it. Each successive coin will need to be raised one more coin thickness, t , than the previous. The thickness of each coin will be t y
N
=
. The work done on each coin will be equal to its increase in gravitational potential energy.T 1 2 N 1 2 N 1 2 N T ( ) (0 2 ) (0 1 2 ) (1 2 ) W W W W mg y mg y mg y mg y y y mg t t Nt mgt N mgy W N N
= + + +
= ∆ + ∆ + + ∆
=
∆ + ∆ + + ∆
=
+ + +
=
+ + +
=
+ + +
The sum of an arithmetic series is: 1 2 n n t t S
=
n
+
For the series (1 + 2 + ··· + N ): 1
+
N
Substituting in the original equation to find the amount of work that must be done on the last coin: T T 1 2 1 2 mgy N W N N N W mgy
+
=
+
=
(b) The energy stored is equal to the work done, therefore:
g 1 2 N E mgy
+
∆ =
Applying Inquiry Skills
7. The units for gravitational potential energy:
( )
2 2 2 m kg m s kg m /s mg y∆ =
= ⋅
The units for work:
( )( )
( )
2 2 2 N m m kg m s kg m /s F d∆ =
=
= ⋅
The units for kinetic energy:
( )
2 2 2 2 1 m kg 2 s kg m /s mv=
= ⋅
Therefore, all three units are the same.
Making Connections
8. ∆ E g = 6.1 109 J (a) ∆ y = ?
Assume 920 students in the school. Assume an average mass of 70.0 kg per student. m = 70.0 kg 920 = 6.44 104 kg g g 9 4 4 6.1 10 J (6.44 10 kg)(9.80 N/kg) 9.472 10 m E mg y E y mg y
∆ =
∆
∆
∆ =
×
=
×
∆ =
×
The energy from one barrel of oil could raise the students 9.472 104 m above ground level. (b) There are 158.987 L in a barrel of oil
9 7 6.1 10 J 3.8 10 J/L 158.987 L
×
=
×
PRACTICE
(Page 193)Understanding Concepts
9. (a) The Sun’s radiant energy is converted to thermal energy in the water, which is then converted into gravitational
potential energy as the water rises. The gravitatio nal energy converts into kinetic energy as the water falls and t urns the turbine. The kinetic energy of the turbine is converted into electrical energy by the generator.
(b) A run-of-the-river generating station does not dam the water to create a large vertical drop in a short area, but rather uses the natural drop of the land over a certain distance, and diverts part of the water to flow down this path.
Making Connections
(c) The main difference students will find is that most run-of-the-river generating stations in Canada are much smaller. (d) (i) rainfall, rivers, and glacier/mountain snow melting
(ii) Bhutan relies heavily on its environment for exports from logging and energy.
(e) possible points: cost, limited suitable locations, low population density, long-term effects
Section 4.3 Questions
(Page 194)Understanding Concepts
1. As the construction worker raises the wood, the wood’s gravitational energy increases. 2. m = 63 kg ∆ y = 3.4 m (a) On Earth, g = 9.80 N/kg ∆ E g = ? g 3 g (63 kg)(9.80 N/kg)(3.4 m) 2.1 10 J E mg y E
∆ =
∆
=
∆ =
×
The astronaut’s gravitational potential energy is 2.1 103 J. (b) On the Moon, g = 1.6 N/kg
∆
E g = ? 2 (63 kg)(1.6 N/kg)(3.4 m) 3.4 10 J g g E mg y E∆ =
∆
=
∆ =
×
The astronaut’s gravitational potential energy is 3.4 102 J. 3. m = 125 g = 0.125 kg
∆
y = 3.50 m(a)
∆
E g = ? (of the pear relative to the ground) The pear on the branch:g g (0.125 kg)(9.80 N/kg)(3.50 m) 4.29 J E mg y E
∆ =
∆
=
∆ =
The pear at ground level: g g (0.125 kg)(9.80 N/kg)(0.00 m) 0.00 J E mg y E
∆ =
∆
=
∆ =
(b)
∆
E g = ? (of the pear relative to the branch) The pear on the branch:g g (0.125 kg)(9.80 N/kg)(0.00 m) 0.00 J E mg y E
∆ =
∆
=
∆ =
The pear on the ground: g g (0.125 kg)(9.80 N/kg)( 3.50 m) 4.29 J E mg y E
∆ =
∆
=
−
∆ = −
The gravitational potential energy of the pear on the branch relative to the branch is 0.00 J. The gravitational potential energy of the pear on the ground relative to the branch is –4.29 J.
4. m = 0.15 kg
∆
E g = 22 J∆
y = ? g g 22 J (0.15 kg)(9.80 N/kg) 15 m E mg y E y mg y∆ =
∆
∆
∆ =
=
∆ =
The ball’s maximum height is 15 m above the point where it was hit. 5. Let the subscript g represent gravity, and W represent the weightlifter.
m = 15 kg
∆
y = 66 cm = 0.66 (a) W g = ? g g ( cos ) ( cos ) (15 kg)( 9.80 N/m)(cos180 )(0.66 m) 97 J W F d mg d W θ θ=
∆
=
∆
=
−
°
= −
The amount of work done by gravity on the mass is –97 J. (b) W W = ? W W ( cos ) ( cos ) (15 kg)(9.80 N/kg)(cos 0 )(0.66 m) 97 J W F d mg d W θ θ
=
∆
=
∆
=
°
=
The amount of work done by the weightlifter on the mass is 97 J. (c)
∆
E g = ? g g (15 kg)(9.80 N/kg)(0.66 m) 97 J E mg y E∆ =
∆
=
∆ =
Applying Inquiry Skills
6.Making Connections
7. (a) V = 32.8 km3 = 3.28 × 1010 m3∆
y = 23.1 m ρ = 1.00 103 kg/m3∆
E g = ?First we must determine the mass of the water:
3 3 10 3 13 (1.00 10 kg/m )(3.28 10 m ) 3.28 10 kg m V m ρ
=
=
×
×
=
×
To calculate the gravitational potential energy: g 13 15 g (3.28 10 kg)(9.80 N/kg)(23.1 m) 7.43 10 J E mg y E
∆ =
∆
=
×
∆ =
×
The gravitational potential energy of the lake relative to the turbines is 7.43 1015 J.
(b) This value is approximately
15 15 7.43 10 J 6.52 1.14 10 J
×
=
×
times the annual energy output of the Chukha plant in Bhutan.4.4 THE LAW OF CONSERVATION OF ENERGY
PRACTICE
(Page 197)Understanding Concepts
1. The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in kinetic energy. This can only be done with negative work.
2. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity.
3.
∆
y = 59.4 m vi = 0.0 m/s vf = ?Using conservation of energy, we will have no kinetic energy at the top, and no gravitational energy at the bottom. T1 T2 2 1 f f 1 2 f 1 2 2 2(9.80 m/s )(59.4 m) 34.1 m/s E E mgy mv v gy v
=
=
/
/
=
=
=
To convert to km/h: m 3600 s 1 km 34.1 123km/h s h 1000 m
=
The roller coaster reaches a maximum speed of 123 km/h at the bottom of the hill. 4. vi = v1 = 9.7 m/s
∆
y = 4.2 m vf = v2= ?We will use y = 0 at the point of contact on the hill. T1 T2 2 1 2 1 1 2 2 2 2 2 1 1 2 2 2 2 2 1 1 2 1 2 2 2 2 2 E E mv mgy mv mgy v gy v gy v v gy gy
=
+
=
+
+
= +
= +
−
Since y2 = 0, 2 gy2 = 0, therefore: 2 2 1 1 2 2 2 2 (9.7 m/s) 2(9.80 m/s )( 4.2 m) 13 m/s v v gy v
=
+
=
+
=
The skier’s speed upon touching the hillside is 13 m/s. 5.
∆
y = 4.4 102 m v2 = 93 m/s v1 = ? T1 T2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 1 2 2 1 2 2 1 1 1 2 2 2 2 2 2 2 ( ) (93 m/s) 2(9.8 m/s )(0 440 m) 5.0 m/s E E mv mgy mv mgy v gy v gy v v gy gy v v g y y v=
+
=
+
+
= +
= +
−
=
+
−
=
+
−
=
The speed of the water at the top of the waterfall is 5.0 m/s. 6. v1 = 9.7 m/s
∆
y = 4.7 m v2 = ?T1 T2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 1 1 2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 2 2 ( ) (9.7 m/s) 2(9.8 m/s )(0 4.7 m) 1.4 m/s E E mv mgy mv mgy v gy v gy v v gy gy v v g y y v
=
+
=
+
+
= +
= +
−
=
+
−
=
+
−
=
The cyclist crests the hill at a speed of 1.4 m/s. 7. v2 = ?
First determine how high the pendulum is vertically raised:
2 2 2 1 2 2 1 1 24.5 cm 85.5 cm 85.5 cm 24.5 cm 81.915 cm h h h
+
=
=
−
=
1 2 2 1 2 85.5 cm 85.5 cm 85.5 cm 81.915 cm 3.585 cm h h h h h+ =
=
−
=
−
=
Using the value h2 = 3.585 cm, or 0.03585 m, calculate the maximum speed: T1 T2 2 2 1 1 2 2 1 2 1 2 2 2 2 1 2 2 1 2 2 1 1 2 2 since 0 2 2 2 2 2 ( ) 2(9.80 m/s )(0.03585 m 0) 0.838 m/s E E mv mgy mv mgy v gy v gy v gy gy v g y y v
=
+
=
+
=
= +
=
−
=
−
=
−
=
Applying Inquiry Skills
8.∆
∆∆
∆
y (m)∆∆
∆
∆
E g (J) E K (J) E T (J)8.00 392 0.00 392 6.00 294 98.0 392 4.00 196 196 392 2.00 98.0 294 392 0.00 0.00 392 392
Making Connections
9. A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released, the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.
PRACTICE
(Page 200)Understanding Concepts
10. (a) The energy supplied becomes sound and thermal energy through friction.
(b) The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy. 11. F K = 67 N
∆
d = 3.5 m (a) W = ? 2 ( cos ) (67 N)(cos180 )(3.5 m) 2.3 10 J W F d W θ=
∆
=
°
= − ×
The amount of work done by friction is –2.3 102 J. (b) E th = ? th K 2 th (67 N)(3.5 m) 2.3 10 J E F d E
= ∆
=
=
×
The amount of thermal energy produced is 2.3 102 J. 12. E th = 0.620 J
F K = 0.83 N
The 0.620 J of energy comes from the work done by friction, therefore: th K th K 0.620 J 0.83 N 0.75 m E F d E d F d
= ∆
∆ =
=
∆ =
The plate slides 0.75 m. 13. m = 22.0 kg F = 98 N F K = 87 N vi = 0.0 m/s
∆
d = 1.2 m vf = ? th k 2 K 2 2 K 2 K 2 2 1 ( cos ) 2 ( cos ) 1 2 ( cos ) 0.5 (98 N)(cos0 )(1.2 m) 87 N(1.2 m) 0.5(22.0 kg) 1.1 m/s W E E F d F d mv F d F d v m F d F d v m v θ θ θ=
+
∆ = ∆ +
∆ − ∆
=
∆ − ∆
=
°
−
=
=
The speed of the cabinet after it moves 1.2 m is 1.1 m/s. 14. m = 0.057 kg
∆
d = 25 cm = 0.25 m vf = 5.7 cm/s = 5.7 10 –2 m/s F K = 0.15 N vi = ? K1 th K2 2 2 1 K 2 2 K 2 2 1 2 K 2 1 2 2 1 1 1 2 2 1 2 1 2 1 2 0.5 1 (0.15 N)(0.25 m) (0.057 kg)(5.7 10 m/s) 2 0.5(0.057 kg) 1.1 m/s E E E mv F d mv F d mv v m F d mv v m v −=
+
= ∆ +
∆ +
=
∆ +
=
+
×
=
=
The initial speed of the pen is 1.1 m/s.
Applying Inquiry Skills
15. (a) The law of conservation of energy could be verified by observing how close to its original position the pendulum would return after a swing.
(b) Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the moving pendulum. Some error may be observed if the string is somewhat elasti c.
Making Connections
16. The oil circulation system is an attempt to minimize the frictional forces between the moving parts of the engine, and thereby reduce the loss (and damage) due to the thermal energy caused by friction. The water (coolant) circulation is used to absorb thermal energy from the engine and dissipate it rapidly into the air to prevent damage from overheating.
Section 4.4 Questions
(Pages 201–202)Understanding Concepts
1. Roller coasters are gravity rides that have an initial input of gravitational potential energy that is converted to kinetic (and back into gravitational) energy throughout the ride. T o give them this initial energy, they must be pulled up the largest hill
at the beginning. 2. m = 0.052 kg
∆
y = 11 cm = 0.11 m y = 0 (a)∆
E g = ? g 2 g (0.052 kg)(9.80 m/s )(0.11 m) 0.056 J E mg y E∆ =
∆
=
∆ =
The initial gravitational potential energy of the egg’s contents is 0.056 J. (b)
∆
E g = ? g 2 g (0.052 kg)(9.80 m/s )(0.0 m) 0.0 J E mg y E∆ =
∆
=
∆ =
The final gravitational potential energy of the egg’s contents is 0.0 J. (c)
∆
E g = 0.0 – 0.056 = –0.056 JThe change in gravitational potential energy as the egg’s contents fall is –0.056 J. (d) E K = ?
v = ?
The kinetic energy will be equal to the loss of gravitational potential, so E K = 0.056 J. 2 K K 1 2 2 2(0.056 J) 0.052 kg 1.5 m/s E mv E v m v
=
=
=
=
The kinetic energy is 0.056 J. The speed of the egg’s contents just before hitting the pan is 1.5 m/s. 3.
∆
y = 1.2 mvf = 9.9 m/s vi = ?
T1 T2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 1 2 2 1 2 2 1 1 1 2 2 2 2 2 2 2 ( ) (9.9 m/s) 2(9.8 m/s )(1.2 m 0 m) 11 m/s E E mv mgy mv mgy v gy v gy v v gy gy v v g y y v
=
+
=
+
+
= +
= +
−
=
+
−
=
+
−
=
The initial speed of the ball was 11 m/s. 4. v1= 3.74 m/s
∆
y = 8.74 m m = 7.12 104 kg (a)∆
E g = ? g 4 2 6 g (7.12 10 kg)(9.80 m/s )(8.74 m) 6.10 10 J E mg y E∆ =
∆
=
×
∆ =
×
The gravitational potential energy of the mass of water at the top of the waterfall is 6.10 106 J. (b) v2 = ? T1 T2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 1 1 2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 2 2 ( ) (3.74 m/s) 2(9.80 m/s )(8.74 m 0 m) 13.6 m/s E E mv mgy mv mgy v gy v gy v v gy gy v v g y y v
=
+
=
+
+
= +
= +
−
=
+
−
=
+
−
=
The speed of the water at the bottom of the waterfall is 13.6 m/s. 5. (a)
First, determine the vertical height above the bottom of the swing:
(
)
1 1 1 cos48 3.7 m 3.7 m cos48 2.476 m h h h° =
=
°
=
1 2 2 1 2 3.7 m 3.7 m 3.7 m 2.476 m 1.224 m h h h h h
+ =
=
−
=
−
=
Now we can calculate the acrobat’s speed: T1 T2 2 2 1 1 2 2 1 1 2 2 E E mv mgy mv mgy
=
+
=
+
Since v1 = 0: 2 1 2 2 2 2 1 2 2 1 2 2 2 2 2 2 2 2 ( ) 2(9.80 m/s )(1.224 m 0 m) 4.9 m/s gy v gy v gy gy v g y y v= +
=
−
=
−
=
−
=
The acrobat’s speed at the bottom of the swing is 4.9 m/s.
(b) Due to conservation of energy, the maximum height on the other side is equal to the starting height of the acrobat. 6. m = 55 kg
∆
d = 3.7 m F K = 41.5 N v1 = 65.7 cm/s = 0.657 m/s v2 = 7.19 m/s φ = ? Relatingφ to h: sin 11.7 11.7sin h h φ φ=
=
Using conservation of energy:
(
)
(
)
(
)
T1 T2 K1 g K2 th 2 2 1 2 K 2 2 2 K 1 2 2 2 1 K 2 2 2 1 1 2 2 1 1 2 2 1 ( ) 2 1 (55.0 kg) 7.19 m/s 0.657 m/s (41.5 N)(11.7 m) 2 11.7sin (55.0 kg)(9.80 m/s ) sin 0.30054 17.5 E E E E E E mv mgh mv F d mgh mv F d mv m v v F d h mg φ φ φ=
+ ∆ =
+
+ =
+ ∆
=
+ ∆ −
−
+ ∆
=
−
+
=
=
=
°
The angle is 17.5°
.7. v1 = 0.0 m/s v2 = 6.8 m/s
Since the skateboarder starts even with the height of the centre of the circle, the total vertical drop at the bottom will be equal to the radius of the circle.
r = y1 = ? T1 T2 2 2 1 1 2 2 2 2 1 1 2 2 1 1 2 2 2 2 E E mv mgy mv mgy v gy v gy
=
+
=
+
+
= +
Since v1 = 0, and y2 = 0: 2 1 2 2 2 1 2 2 1 2 2 6.8 m/s 2(9.8 m/s ) 2.4 m gy v v y g y=
=
=
=
The radius of the half-pipe is 2.4 m. 8. m = 55 g = 5.5 10 –2 kg
v1 = 1.9 m/s
∆
d = 54 cm = 0.54 m v2 = 0.0 m/s(a) µ K = ?
First, solve for F K :
T1 T 2 K th 2 K 2 K 2 2 K 1 2 2 (5.5 10 kg )(1.9 m/ s) 2(0.54 m) 0.1838 N E E E E mv F d mv F d F −
=
=
= ∆
=
∆
×
=
=
Now calculate the normal force:
N N 2 2 N 0 0 (5.5 10 kg)(9.80 m/s ) 0.539 N y y F ma F mg F mg F −
Σ =
=
−
=
=
=
×
=
Solve for µK : K K N 0.1838 N 0.539 N 0.34 K F F µ µ=
=
=
(b) µ K =?
First, solve for the acceleration:
(
) ( )
2 2 f i 2 2 f i 2 2 2 2 2 1.9 m/s 0 m/s 2(0.54 m) 3.3426 m/s v v a d v v a d a= + ∆
−
=
∆
−
=
=
Using the FBD, calculate the magnitude of kinetic friction:
K 2 2 K (5.5 10 kg)(3.3426 m/s ) 0.1838 N x x x F ma F ma F −
Σ =
=
=
×
=
Calculate the normal force:
N N 2 2 N 0 0 (5.5 10 kg)(9.80 m/s ) 0.539 N y y F ma F mg F mg F −
Σ =
=
−
=
=
=
×
=
Calculate the coefficient of kinetic friction: K K N K 0.1838 N 0.539 N 0.34 F F µ µ
=
=
=
The coefficient of kinetic friction is 0.34.
(c) The kinetic energy is converted into thermal energy. 9. vi = 85 km/h = 23.61 m/s vf = 0.0 m/s
∆
d = 47 m F K = 7.4 103 N (a) E th = ? th K 3 5 th (7.4 10 N)(47 m) 3.5 10 J E F d E= ∆
=
×
=
×
The amount of thermal energy produced is 3.5 105 J. (b) Before the skid, the thermal energy was kinetic energy.
(c) m = ?
(
)
T1 T2 K th 2 K K 2 3 2 3 1 2 2 2(7.4 10 N )(47 m ) 23.61 m/s 1.2 10 kg E E E E mv F d F d m v m=
=
= ∆
∆
=
×
=
= ×
The mass of the car is 1.2 103 kg. (d) µ K = ?
First calculate the normal force:
N N 3 2 N 0 0 (1.248 10 kg)(9.80 m/s ) 12 228 y y F ma F mg F mg F
Σ =
=
−
=
=
=
×
=
To calculate the coefficient of kinetic friction: K K N 3 K 7.4 10 J 12228 J 0.61 F F µ µ
=
×
=
=
The coefficient of kinetic friction is 0.61. 10. m = 22 kg
∆
d = 2.5 m θ = 44°
F K = 79 N (a) W = ? K 2 ( cos ) (79 N)(cos180 )(2.5 m) 2.0 10 J W F d W θ=
∆
=
°
= − ×
The work done by friction is –2.0 102 J. (b) E K = ?
First, determine the vertical drop: sin44 2.5 2.5sin 44 1.737 m h h h
° =
=
°
=
To calculate the final kinetic energy: T1 T2 g K th K K K K 2 K (9.8 N/kg)(22 kg)(1.737 m) (79 N)(2.5 m) 1.8 10 J E E E E E mg y E F d E mg y F d E
=
∆ =
+
∆ = + ∆
= ∆ − ∆
=
−
= ×
The box’s final kinetic energy is 1.8 102 J.
(c) E th = ? th K 2 th (79 N)(2.5 m) 2.0 10 J E F d E
= ∆
=
=
×
The thermal energy produced is 2.0 102 J.
Applying Inquiry Skills
11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energy produced.
(b) The equation needed would be: T1 T2 g1 g2 th 1 2 th th 1 2 th ( 1 2) E E E E E mgy mgy E
E mgy mgy E mg y y
=
∆ = ∆ +
=
+
=
−
=
−
(c) You could determine the height at each point by standing a known distance from the base of the ride and measure the angle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energy
produced.
Making Connections
12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potential energy is converted into the kinetic energy of the projectile.
4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTION
PRACTICE
(Pages 206–207)Understanding Concepts
1. The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring A would be more difficult to stretch than spring B.
2. The spring would exert a southward force on you. 3. k = 25 N/m (a) x = 16 cm = 0.16 m (25 N/m)(0.16 m) 4.0 N x x F kx F
=
=
=
x = 32 cm = 0.32 m (25 n/m)(0.32 m) 8.0 N x x F kx F
=
=
=
The magnitude of force would be 4.0 N for a stretch of 16 cm, and 8.0 N for a stretch of 32 cm. (b) x = 16 cm = 0.16m (25 N/m)(0.16 m) 4.0 N x x F kx F
= −
= −
= −
x = 32 cm = 0.32 m (25 N/m)(0.32 m) 8.0 N x x F kx F= −
= −
= −
The magnitudes of the forces are 4.0 N and 8.0 N, respectively. 4. k = 3.2 102 N/m x = 2.0 cm = 2.0 10 –2 m 2 2 (3.2 10 N/m)(2.0 10 m) 6.4 N x x F kx F −
= −
=
− ×
×
= −
The magnitude of the force applied by the air is 6.4 N. 5. m = 1.37 kg k = 5.20 102 N/m (a) x = ? 2 0 0 (1.37 kg)(9.80 N/kg) 5.20 10 N/m 0.0258 m y x F ma F mg kx mg mg x k x
Σ =
=
−
=
=
=
=
×
=
The spring stretches 0.0258 m. (b) x = 1.59 cm = 0.0159 m y F
Σ
= ? 2 (5.20 10 N/m)(0.0159 m) (1.37 kg)(9.80 N/kg) 5.16 N y x y F F mg kx mg FΣ = −
= −
=
×
−
Σ = −
The net force on the fish is 5.16 N [down]. (c) x = 2.05 cm = 0.0205 m a = ? 2 (5.20 10 N/m)(0.0205 m) (1.37 kg)(9.80 N/kg) 1.37 kg y x F ma F mg ma kx mg a m
Σ =
−
=
−
=
×
−
=
Applying Inquiry Skills
6. (a)(b) The slope of the line is negative.
Making Connections
7. (a) Mass (kg) Stretch (m) 1.00 0.122 2.00 0.245 3.00 0.368 4.00 0.490 5.00 0.612 6.00 0.735 7.00 0.858 8.00 0.980 (b)(c) The mass value may not be correct if the value of g is different then where it was calibrated. The weight value would be accurate anywhere (even on the Moon).
PRACTICE
(Page 211)Understanding Concepts
8. (a) The graph shown is the force applied by the spring. Since the spring is being stretched to the right (positive x), the force exerted by the spring will be to the left (negative). This is what is shown on the graph.
(b) The force constant is the slope of the graph. Since the equation F x= kx relates the force exerted by the spring, we must change the direction (i.e., the sign) of the force,
( 15 N) 0.40 m 38 N/m x F k x k
−
=
− −
=
=
The force constant of the spring is 38 N/m. (c) x = 35 cm = 0.35 m
The energy stored is the area between the curve and the x-axis.
e e 1 2 1 (0.35 m)(13 N) 2 2.3 J E A bh E
= =
=
=
The elastic potential energy is 2.3 J. 9. k = 9.0 103 N/m (a) x = 1.0 cm = 0.010 m E e = ? 2 e 3 2 e 1 2 1 (9.0 10 N/m)(0.010 m) 2 0.45 J E kx E
=
=
×
=
The elastic potential energy stored by the spring is 0.45 J. (b) x = –2.0 cm = –0.020 m E e = ? 2 e 3 2 e 1 2 1 (9.0 10 N/m)( 0.020 m) 2 1.8 J E kx E
=
=
×
−
=
The elastic potential energy stored in the spring is 1.8 J. 10. m = 7.8 g = 0.0078 kg k = 3.5 102 N/m x = –4.5 cm = –0.045 m (a) E e = ? 2 e 2 2 e 1 2 1 (3.5 10 N/m)( 0.045 kg) 2 0.35 J E kx E
=
=
×
−
=
(b) v = ? T1 T2 e K 2 2 2 2 2 1 1 2 2 (3.5 10 N/m)( 0.045 m) 0.0078 kg 9.5 m/s E E E E kx mv kx v m v
=
=
=
=
×
−
=
=
The speed of the dart as it leaves the toy is 9.5 m/s. 11. m = 3.5 10 –3 kg k = 9.5 N/m (a)
∆
y = 5.7 cm = 0.057 m x = ? T1 T2 e g 2 3 1 2 2 2(3.5 10 kg)(9.8 N/kg)(0.057 m) 9.5 N/m 0.020 m E E E E kx mg y mg y x k x −=
= ∆
=
∆
∆
=
×
=
=
The spring must be compressed by 0.020 m, or 2.0 cm.
(b) If friction was not negligible, the amount of compression would need to be increased. The energy supplied by the compressed spring would now become both gravitational potential and thermal energy. In order to have the same final gravitational energy, the spring would need to be compressed more.
12. m = 0.20 kg k = 55 N/m (a)
∆
y = 1.5 cm = 0.015 m v = ? T1 T2 g K e 2 2 2 2 2 2 1 1 2 2 2 2 2(0.20 kg)(9.8 N/kg)(0.015 m) (55 N/m)(0.015 m) 0.20 kg 0.48 m/s E E E E E mg y mv kx mv mg y kx mg y kx v m v=
∆ =
+
∆ =
+
=
∆ −
∆ −
=
−
=
=
(b) For the fall,
∆
y = x. At maximum fall, there will be no kinetic energy.( )
( ) ( )
T1 T2 g e 2 2 2 2 1 2 2 0 55 2(0.20)(9.8) 0 55 3.92 0 (55 3.92) 0 55 3.92 0 or 0 3.92 55 0.071m or 0 m E E E E mg y kx kx mg y x x x x x x x x x x x=
∆ =
∆ =
−
∆ =
−
=
−
=
−
=
−
=
=
=
=
=
The value x = 0 refers to the moment of release, so the maximum stretch will be 0.071 m. 13. k = 12 N/m
∆
y = 93.0 cm = 0.93 m m = 8.3 10 –3 kg x = 4.0 cm = 0.040 m∆
d = ?Analyze as a projectile motion question. First, determine horizontal speed at launch: T1 T2 e K 2 2 2 2 3 1 1 2 2 (12 N/m)( 0.040 m) 8.3 10 kg 1.521 m/s E E E E kx mv kx v m v −
=
=
=
=
−
=
×
=
Choosing down as positive, the time for the marble to drop 0.93 m vertically is: 2 i 1 ( ) 2 y v t a t
∆ = ∆ + ∆
Since vi = 0: 2 2 1 ( ) 2 2 2(0.93 m) 9.8 m/s 0.4357 s y a t y t a t∆ =
∆
∆
∆ =
=
∆ =
The horizontal distance travelled during this time is:
(1.521 m/s)(0.4357 s) d v t
∆ = ∆
=
∆ =
Applying Inquiry Skills
14. (a) The measurement needed would be the mass of the largest friend. (b) Assuming a largest mass of 115 kg,
3 0 0 (115 kg)(9.80 N/kg) 0.75 m 1.5 10 N/m y x F ma F mg kx mg mg k x k
Σ =
=
−
=
=
=
=
= ×
The approximate force constant is 1.5 103 N/m.
Making Connections
15. m = 2.0 102µ g = 2.0 10 –7 kg∆
y = 65 mm = 0.065 m x = 75 cm = 0.75 m E e = ? T1 T2 e g 7 8 e ( ) ( ) 0.75 m(2.0 10 kg)(9.8 N/kg)(0.065 m) 9.6 10 J E E E x E x mg y E − −=
= ∆
=
∆
=
×
=
×
The initial quantity of elastic potential energy is 9.6 10 –8 J.
PRACTICE
(Pages 214–215)
Understanding Concepts
16. (a) The maximum displacement from the rest position will be at the top and bottom of the bounce. (b) The speed is a maximum at the rest position.
(c) The speed will be zero at the top and the bottom of the bounce.
(d) The acceleration will be a maximum at the top and the bottom of the bounce. (e) The acceleration will be zero at the rest position.
17. T = ? f = ?
(a) number of vibrations = 12 t = 48 s
total time
number of complete vibrations 48 12 4.0 s T T
=
=
=
number of complete vibrations total time 12 48 0.25 Hz f f
=
=
=
Alternatively, you could calculate frequency by using the equation: 1 1 4.0 s 0.25 Hz f T f
=
=
=
The period is 4.0 s, and the frequency is 0.25 Hz. (b) number of vibrations = 210
t = 1 min = 60 s
total time
number of complete vibrations 60 210 0.29 s T T
=
=
=
number of complete vibrations total time 210 60 3.5 Hz f f
=
=
=
The period is 0.29 s, and the frequency is 3.5 Hz. (c) number of vibrations = 2200
t = 5.0 s
3
total time
number of complete vibrations 5.0 2200 2.3 10 s T T −
=
=
=
×
2number of complete vibrations total time 2200 5.0 4.4 10 Hz f f
=
=
=
×
The period is 2.3 10 –3 s, and the frequency is 4.4 102 Hz. 18. m = 0.25 kg
A = 8.5 cm
k = 1.4 102 N/m
(a) d = ?
During each cycle, the mass moves 4 amplitudes, or 4 A. In 5 cycles, the mass will move:
2 5 4 5 4(8.5 cm) 1.7 10 cm d A d
= ×
= ×
=
×
The mass moves 1.7 102 cm in the first five cycles. (b) T = ? 2 2 0.25 kg 2 1.4 10 N/m m T k π π
=
=
×
19. m = 0.10 kg f = 2.5 Hz k = ? 2 2 2 2 1 2 2 4 4 (2.5 Hz) (0.10 kg) 25 N/m k f m k f m k f m k π π π π
=
=
=
=
=
The force constant of the spring is 25 N/m. 20. k = 1.4 102 N/m T = 0.85 s m = ? 2 2 2 2 2 2 2 4 (0.85 s) (1.4 10 N/m) 4 2.6 kg m T k T m k T k m m π π π π
=
=
=
×
=
=
The mass would have to be 2.6 kg.
Applying Inquiry Skills
21. Examining the base SI units for each:
2 2 m m s s s x a x a
=
=
=
2 2 kg N m kg m = kg m s s s m k m k=
⋅
⋅
=
=
Therefore, they are dimensionally equivalent.
Making Connections
22. number of vibrations = 6.0 t = 8.0 s
(a) k = ?
First we must calculate the frequency:
number of complete vibrations total time 6.0 8.0 s 0.75 Hz f f
=
=
=
Using a mass of 75 kg: 2 2 2 2 3 1 2 2 4 4 (0.75 Hz) (75 kg) 1.7 10 N/m k f m k f m k f m k π π π π
=
=
=
=
= ×
The force constant is 1.7 103 N/m.
(b) No, you are not undergoing SHM. When you leave the trampoline, there is a period of time when it is not exerting any force on you. SHM requires that the force be proportional to the displacement.
PRACTICE
(Pages 217–218)Understanding Concepts
23. (a) The speed is zero at lengths of 12 cm and 38 cm.
(b) The maximum speed will be at the rest position. The rest position will be halfway between the minimum and maximum extensions: 12 38 25 cm 2
+
=
(c) The amplitude is 38 – 25 = 13 cm. 24. E e = 5.64 J m = 0.128 kg k = 244 N/m (a) x = ?The maximum energy is constant, and all elastic potential at either end of the system, at the maximum amplitude. 2 e e 1 2 2 2(5.64 J) 244 N/m 0.215 m E kx E x k x
=
=
=
=
The amplitude of the vibration is 0.215 m. (b) v = ?
Approach 1: All of the energy will be kinetic as it passes through the rest position. 2 K K 2 2(5.64 J) 0.128 kg 9.39 m/s E mv E v m v
=
=
=
=
Approach 2: All of the energy stored in the spring at maximum compression will be converted to kinetic energy. T1 T2 2 2 2 2 1 1 2 2 (244 N/m)(0.215 m) 0.128 kg 9.39 m/s E E kx mv kx v m v
=
=
=
=
=
The speed is 9.39 m/s regardless of the approach used. (c) x2= 15.5 cm = 0.155 m v = ?
(
)
(
)
(
)
T1 T2 2 2 2 max 2 2 2 2 max 2 2 2 max 2 2 2 1 1 1 2 2 2 ( ) (244 N/m) 0.215 m 0.155 m 0.128 kg 6.51 m/s E E kx mv kx kx kx v m k x x v m v=
=
+
−
=
−
=
−
=
=
The speed of the mass is 6.51 m/s. 25. x = 0.18 m
m = 58 g = 0.058 kg k = 36 N/m
(a) E e = ? v = ?
Maximum energy during maximum stretch/compression of the spring: 2 max 2 max 1 2 1 (36 N/m)(0.18 m) 2 0.58 J E kx E
=
=
=
This will all be kinetic energy at the rest position. 2 K K 1 2 2 2(0.58 J) 0.058 kg 4.5 m/s E mv E v m v
=
=
=
=
(b) x = ?
Double the energy would be: max e 2 2 max 2 1 2 2 (36 N/m)(0.18 m) 1.1664 J E E kx E
′ = ×
= ×
=
′ =
This is all stored as elastic potential energy at the full amplitude. 2 e e 1 2 2 2(1.1664 J) 36 N/m 0.25 m E kx E x k x
=
=
=
=
The amplitude of vibration required would be 0.25 m. (c) v = ? 2 K K 1 2 2 2(1.1664 J) 0.058 kg 6.3 m/s E mv E v m v
=
=
=
=
The maximum speed of the mass is 6.3 m/s.
26. For the maximum speed, all of the elastic energy will be converted to kinetic: T1 T2 2 2 2 1 1 2 2 E E kx mv kx v m k v x m
=
=
=
=
For a SHM system, x = A, therefore: 1 2 2 k f m k f m π π
=
=
Substituting above: (2 ) 2 k v x m A f v fA π π=
=
=
Applying Inquiry Skills
27. (a) 2 2 2 2 N m (m m ) kg k (A x ) m
−
=
−
2 2 2 2 kg m m s (m ) kg m s m/s⋅
⋅
=
=
=
The dimensions are m/s.
(b) This expression is to calculate the speed of an object at a location during SHM.
Making Connections
28. (a) Tuning fork prongs have slow damping to produce a long tone.
(b) The voltmeter needle has fast damping to stabilize the reading quickly. (c) A guitar string has slow damping to play the note as long as possible.
(d) Saloon doors have medium damping to keep the doors from swinging too much, but they still swing back and forth. (e) The string of the bow has fast damping to prevent dangerous vibration.
Section 4.5 Questions
(Pages 218–219)Understanding Concepts
1. When the two students pull on either end of the spring, it will not stretch as much as when it is pulled by both students while attached to the wall. When they both pull on it from the same side, the wall pulls back with equal force. When they pull from opposite ends, the force will be hal f as much, and the stretch will also be half as much.
2. The elastic potential energy is the same when the spring is stretched or compressed 2.0 cm. The amount of energy stored only depends on the magnitude of the distortion, not the direction.
3. Harmonic means that it is regularly repeated, symmetrical motion. 4. (a) Period is inversely proportional to the frequency, T 1
f
∝
.(b) The acceleration is directly proportional to the displacement, a
∝
x.(c) The period is inversely proportional to square root of the force constant, T 1 k
∝
.(d) The maximum speed is directly proportional to the amplitude, v
∝
A. 5. m = 62 kg k = 2.4 103 N/m x = ? 3 0 6 0 6 6 (62 kg)(9.8 N/kg) 6(2.4 10 N/m) 0.042 m y x F ma F mg kx mg mg x k xΣ =
=
−
=
=
=
=
×
=
6. k = 78 N/m x = 2.3 cm = 0.023 m F x = ? (78 N/m)(0.023 m) 1.8 N x x F kx F
=
=
=
The magnitude of force is 1.8 N. 7. x1 = 1.85 cm
F x1 = 85.5 N
x2 = 4.95 cm = 0.0495 m F x2 = ?
First we must calculate k : 1 1 1 1 85.5 N 0.0185 m 4621.6 N/m x x F kx F k x k
=
=
=
=
Solve for x2: 2 2 2 (4621.6 N/m)(0.0495 m) 229 N x x F kx F=
=
=
The force required is 229 N. 8. m = 97 kg
k = 2.2 103 N/m
a = 0.45 m/s2 x = ?
Ignoring friction to calculate the applied force:
A 2 A (97 kg)(0.45 m/s ) 43.65 N x F ma F ma F
Σ =
=
=
=
This force is the force exerted on the spring:
3 43.65 N 2.2 10 N/m 0.020 m x x F kx F x k x
=
=
=
×
=
The spring stretches 0.020 m, or 2.0 10 –2 m. 9. m = 289 g = 0.289 kg k = 18.7 N/m (a) x = 10.0 cm = 0.100 m y F
Σ =
? a = ?To calculate the net force: (18.7 N/m)(0.100 m) (0.289 kg)(9.80 N/kg) 0.962 N y x y F F mg kx mg F
Σ = −
= −
=
−
Σ = −
To calculate acceleration: 2 0.962 N 0.289 kg 3.33 m/s y y F ma F a m aΣ =
Σ
=
−
=
= −
The net force is 0.962 N [down], and the acceleration is 3.33 m/s2 [down]. (b) x = ? 0 0 (0.289 kg)(9.80 N/kg) 18.7 N/m 0.151 m y x F ma F mg kx mg mg x k x
Σ =
=
−
=
=
=
=
=
The spring will be stretched by 0.151 m. 10. m = 64.5 kg ∆ y = 48.0 m –12.5 m = 35.5 m k = 65.5 N/m x = 35.5 m – 10.1 m = 25.4 m. v = ? g e K 2 2 2 2 2 2 1 1 2 2 2 2 2(64.5 kg)(9.80 N/kg)(35.5 m) (65.5 N/m)(25.4 m) 64.5 kg 6.37 m/s E E E mg y kx mv mv mg y kx mg y kx v m v
∆ = +
∆ =
+
=
∆ −
∆ −
=
−
=
=
The jumper’s speed at a height of 12.5 m above the water is 6.37 m/s. 11. F x= 8.6 N x = 9.4 cm = 0.094 m (a) k = ? 8.6 N 0.094 m 91 N/m x x F kx F k x k
