49 worked answers · short steps first, explanations when you need them
Chapter Examples, Activities, and Review Questions
Textbook pages 116–139
Question Think It Over 1
What determines a child’s speed at the bottom of a slide? Do two children of different masses reach the same speed? Which slide gives the greatest speed?
Steps only
If friction is ignored, gravitational potential energy mgh becomes kinetic energy ½mv². Thus v = √(2gh): the speed depends on vertical drop h, not mass or slide shape. A slide with the greatest vertical drop gives the greatest bottom speed. In real life, friction may change the result.
Answer
If friction is ignored, gravitational potential energy mgh becomes kinetic energy ½mv². Thus v = √(2gh): the speed depends on vertical drop h, not mass or slide shape. A slide with the greatest vertical drop gives the greatest bottom speed. In real life, friction may change the result.
Question Example 7.1
When is a girl’s work positive or negative as she lifts and lowers a dumbbell?
Steps only
While lifting, her upward force and the dumbbell’s displacement are both upward, so she does positive work. While lowering it slowly, her upward supporting force is opposite to the downward displacement, so her work is negative.
Answer
While lifting, her upward force and the dumbbell’s displacement are both upward, so she does positive work. While lowering it slowly, her upward supporting force is opposite to the downward displacement, so her work is negative.
Question Example 7.2
A goalkeeper exerts 200 N to stop a ball while her hand moves 15 cm backward. Find her work on the ball.
Steps only
Displacement in the direction of the force is −0.15 m. Work = force × displacement = 200 × (−0.15) = −30 J. The negative sign means the force removes energy from the ball.
Answer
Displacement in the direction of the force is −0.15 m. Work = force × displacement = 200 × (−0.15) = −30 J. The negative sign means the force removes energy from the ball.
Question Pause and Ponder 1
Is a weightlifter doing work on a barbell while holding it steady?
Steps only
No mechanical work is done on the barbell because it has no displacement, even though the lifter uses energy to hold it.
Answer
No mechanical work is done on the barbell because it has no displacement, even though the lifter uses energy to hold it.
Question Pause and Ponder 2
Is friction’s work on a stack of coins sliding over a rough surface positive, negative, or zero?
Steps only
Negative. Friction acts opposite the stack’s motion, so it reduces the stack’s kinetic energy.
Answer
Negative. Friction acts opposite the stack’s motion, so it reduces the stack’s kinetic energy.
Question Example 7.3
In a carrom shot, what work and energy changes happen when the striker hits the white coin and it hits the black coin?
Steps only
The striker does positive work on the white coin, giving it kinetic energy. The white coin then does positive work on the black coin and transfers energy to it. The struck coin pushes back on the striker/white coin, doing negative work on it and reducing its energy.
Answer
The striker does positive work on the white coin, giving it kinetic energy. The white coin then does positive work on the black coin and transfers energy to it. The struck coin pushes back on the striker/white coin, doing negative work on it and reducing its energy.
Question Pause and Ponder 3
When you pedal a bicycle on a flat road, into what forms does food energy change?
Steps only
Chemical energy in food changes into mechanical energy of the moving bicycle and rider. Some also changes into heat and sound because of muscle effort and friction.
Answer
Chemical energy in food changes into mechanical energy of the moving bicycle and rider. Some also changes into heat and sound because of muscle effort and friction.
Question Pause and Ponder 4
Two objects of masses m and 4m have equal kinetic energy. Find the ratio of their speeds vA:vB.
Does kinetic energy change with position if an object moves at constant velocity?
Steps only
No. Kinetic energy is ½mv². If mass and speed stay constant, kinetic energy stays constant wherever the object is.
Answer
No. Kinetic energy is ½mv². If mass and speed stay constant, kinetic energy stays constant wherever the object is.
Question Example 7.4
What happens to kinetic energy if a vehicle’s speed doubles?
Steps only
Kinetic energy is proportional to speed squared. At speed 2v, K = ½m(2v)² = 4(½mv²). It becomes four times as large.
Answer
Kinetic energy is proportional to speed squared. At speed 2v, K = ½m(2v)² = 4(½mv²). It becomes four times as large.
Question Example 7.5
Find the kinetic energy of a 0.2 kg cricket ball moving at 154.8 km h⁻¹.
Steps only
Convert speed: 154.8 km h⁻¹ = 43 m s⁻¹. K = ½mv² = ½ × 0.2 × 43² = 184.9 J (about 185 J).
Answer
Convert speed: 154.8 km h⁻¹ = 43 m s⁻¹. K = ½mv² = ½ × 0.2 × 43² = 184.9 J (about 185 J).
Question Example 7.6
A 15,000 kg jet is stopped by a 367,500 N wire over 100 m. Find its speed before the wire caught it.
Steps only
The wire does negative work: W = −367,500 × 100 = −36,750,000 J. This equals the change in kinetic energy: 0 − ½mv² = W. So ½ × 15,000 × v² = 36,750,000; v² = 4900 and v = 70 m s⁻¹ (252 km h⁻¹).
Answer
The wire does negative work: W = −367,500 × 100 = −36,750,000 J. This equals the change in kinetic energy: 0 − ½mv² = W. So ½ × 15,000 × v² = 36,750,000; v² = 4900 and v = 70 m s⁻¹ (252 km h⁻¹).
Question Example 7.7
A 0.2 kg ball reaches 10 m height. Find its potential energy using g = 10 m s⁻².
Steps only
U = mgh = 0.2 × 10 × 10 = 20 J.
Answer
U = mgh = 0.2 × 10 × 10 = 20 J.
Question Pause and Ponder 6
Does potential energy change when an object moves horizontally at constant height? What if it is raised?
Steps only
At the same height, gravitational potential energy mgh does not change. When raised by a height, it increases by mg times the height gained.
Answer
At the same height, gravitational potential energy mgh does not change. When raised by a height, it increases by mg times the height gained.
Question Activity 7.2
What energy changes occur at the two ends and the bottom of a swinging pendulum?
Steps only
At an extreme point, speed is momentarily zero and gravitational potential energy is greatest. At the bottom, potential energy is lowest and kinetic energy is greatest. On the other side, kinetic energy changes back into potential energy. Air resistance and pivot friction gradually transfer mechanical energy into heat and sound.
Answer
At an extreme point, speed is momentarily zero and gravitational potential energy is greatest. At the bottom, potential energy is lowest and kinetic energy is greatest. On the other side, kinetic energy changes back into potential energy. Air resistance and pivot friction gradually transfer mechanical energy into heat and sound.
Question Example 7.8
Find the speed at the bottom of a frictionless slide of vertical height h.
Steps only
At top: potential energy = mgh. At bottom: kinetic energy = ½mv². Equate them: mgh = ½mv², so v = √(2gh). Mass cancels; only the height matters.
Answer
At top: potential energy = mgh. At bottom: kinetic energy = ½mv². Equate them: mgh = ½mv², so v = √(2gh). Mass cancels; only the height matters.
Question Example 7.9
A 10,000 kg truck travels at 72 km h⁻¹ and stops on a 30° escape ramp. Sand pushes back with 50,000 N. Find the minimum ramp length; the truck gains 1 m height for every 2 m along the ramp.
Steps only
72 km h⁻¹ = 20 m s⁻¹. Initial kinetic energy = ½ × 10,000 × 20² = 2,000,000 J. If ramp length is d, height gained = d/2, so gain in potential energy = mgh = 50,000d J. Sand removes another 50,000d J of energy. Thus 2,000,000 = 100,000d, giving d = 20 m.
Answer
72 km h⁻¹ = 20 m s⁻¹. Initial kinetic energy = ½ × 10,000 × 20² = 2,000,000 J. If ramp length is d, height gained = d/2, so gain in potential energy = mgh = 50,000d J. Sand removes another 50,000d J of energy. Thus 2,000,000 = 100,000d, giving d = 20 m.
Question Pause and Ponder 7
For the falling ball in Fig. 7.19, show its mechanical energy just before hitting the ground is mgh.
Steps only
Ignoring air resistance, mechanical energy is conserved. Just before hitting the ground, height is nearly zero, so potential energy is nearly zero. Its kinetic energy has become mgh. Total = 0 + mgh = mgh.
Answer
Ignoring air resistance, mechanical energy is conserved. Just before hitting the ground, height is nearly zero, so potential energy is nearly zero. Its kinetic energy has become mgh. Total = 0 + mgh = mgh.
Question Pause and Ponder 8
How do kinetic and potential energy change at points A, B, C on the ball-track figure? Why are later peaks lower?
Steps only
At a high point A, potential energy is high and kinetic energy low. As the ball descends toward B, potential energy changes to kinetic energy; at the low point, speed and kinetic energy are greatest. As it climbs toward C, kinetic changes back to potential. Later peaks are lower because friction and air resistance transfer some mechanical energy into heat and sound.
Answer
At a high point A, potential energy is high and kinetic energy low. As the ball descends toward B, potential energy changes to kinetic energy; at the low point, speed and kinetic energy are greatest. As it climbs toward C, kinetic changes back to potential. Later peaks are lower because friction and air resistance transfer some mechanical energy into heat and sound.
Question Example 7.10
A weightlifter raises 75 kg by 2 m in 5 s. Find power using g = 10 m s⁻².
Steps only
Work = mgh = 75 × 10 × 2 = 1500 J. Power = work/time = 1500/5 = 300 W.
Answer
Work = mgh = 75 × 10 × 2 = 1500 J. Power = work/time = 1500/5 = 300 W.
Question Example 7.11
A 1000 kg car accelerates from rest to 72 km h⁻¹ in 10 s. Find average engine power, ignoring losses.
Steps only
72 km h⁻¹ = 20 m s⁻¹. Work equals kinetic-energy gain = ½ × 1000 × 20² = 200,000 J. Average power = 200,000/10 = 20,000 W = 20 kW.
Answer
72 km h⁻¹ = 20 m s⁻¹. Work equals kinetic-energy gain = ½ × 1000 × 20² = 200,000 J. Average power = 200,000/10 = 20,000 W = 20 kW.
Question Activity 7.3
What happens to the effort needed to pull a cart up a ramp as the ramp becomes longer and less steep?
Steps only
The required force becomes smaller, but the cart must travel a longer distance. Ignoring friction, the work stays about the same because force decreases while distance increases.
Answer
The required force becomes smaller, but the cart must travel a longer distance. Ignoring friction, the work stays about the same because force decreases while distance increases.
Question Example 7.12
A ramp rises 30 cm and has horizontal width 40 cm. Find its mechanical advantage.
Steps only
Ramp length = √(30² + 40²) = 50 cm. For an ideal inclined plane, mechanical advantage = length/height = 50/30 ≈ 1.67.
Answer
Ramp length = √(30² + 40²) = 50 cm. For an ideal inclined plane, mechanical advantage = length/height = 50/30 ≈ 1.67.
Question Pause and Ponder 9
Why do roads on hills wind around in gentle slopes rather than go straight up?
Steps only
The winding road is a longer, gentler inclined plane. It lets a vehicle gain the same height using a smaller force over a longer distance.
Answer
The winding road is a longer, gentler inclined plane. It lets a vehicle gain the same height using a smaller force over a longer distance.
Question Pause and Ponder 10
Why is an inclined ladder easier to climb than a vertical ladder?
Steps only
The inclined ladder is a longer path to the same height, so less force is needed at each step. The distance is greater; ideal work is not reduced.
Answer
The inclined ladder is a longer path to the same height, so less force is needed at each step. The distance is greater; ideal work is not reduced.
Question Activity 7.4
How can a ruler balanced on a pencil help lift a heavier stapler with a lighter eraser?
Steps only
The pencil is the fulcrum. Put the heavier load close to the fulcrum and apply effort farther away. A longer effort arm lets a smaller effort balance a larger load.
Answer
The pencil is the fulcrum. Put the heavier load close to the fulcrum and apply effort farther away. A longer effort arm lets a smaller effort balance a larger load.
Question Activity 7.5
What relation balances a beam lever with coins on its two sides?
Steps only
A balanced lever has equal turning effects: effort × effort arm = load × load arm. In the coin activity, the beam balances when n₁L₁ = n₂L₂.
Answer
A balanced lever has equal turning effects: effort × effort arm = load × load arm. In the coin activity, the beam balances when n₁L₁ = n₂L₂.
Question Example 7.13
On a seesaw, seats A and E are 2 m from fulcrum C and B and D are 1 m away. Where should 15 kg and 30 kg children sit to balance?
Steps only
Balance requires mass × distance to be equal on both sides. Put 15 kg at A (2 m): 15 × 2 = 30. Put 30 kg at D (1 m): 30 × 1 = 30. They balance.
Answer
Balance requires mass × distance to be equal on both sides. Put 15 kg at A (2 m): 15 × 2 = 30. Put 30 kg at D (1 m): 30 × 1 = 30. They balance.
Question Pause and Ponder 11
Why is it easier to open a can lid with a spoon used as a lever?
Steps only
The spoon’s edge acts as a fulcrum. The long handle gives a long effort arm, so a small hand force can create a larger lifting force on the lid.
Answer
The spoon’s edge acts as a fulcrum. The long handle gives a long effort arm, so a small hand force can create a larger lifting force on the lid.
Question Pause and Ponder 12
Why do you place a hard object near the scissors’ fulcrum?
Steps only
Closer to the fulcrum, the load arm is shorter. This increases the mechanical advantage, so the blades apply a larger force to cut the hard object.
Answer
Closer to the fulcrum, the load arm is shorter. This increases the mechanical advantage, so the blades apply a larger force to cut the hard object.
Question Pause and Ponder 13
Why do real machines slow down and stop instead of working forever without fuel?
Steps only
Friction and air resistance transfer useful mechanical energy into heat and sound. Machines cannot create energy, so without more input their useful motion eventually stops.
Answer
Friction and air resistance transfer useful mechanical energy into heat and sound. Machines cannot create energy, so without more input their useful motion eventually stops.
Question Revise 1
State whether each statement is true or false: (i) force without movement does work; (ii) lifting a bucket does positive work; (iii) work and energy use joules; (iv) a motionless stretched band has kinetic energy; (v) energy changes form.
(i) Force × displacement in the force direction. (ii) 1 newton. (iii) ½mv². (iv) mgh. (v) Rate at which work is done.
Answer
(i) Force × displacement in the force direction.
(ii) 1 newton.
(iii) ½mv².
(iv) mgh.
(v) Rate at which work is done.
Question Revise 3
At the highest point of an upward-thrown ball, which statements are correct: force zero, acceleration zero, kinetic energy zero, potential energy maximum?
Steps only
(iii) and (iv) are correct. Its speed is momentarily zero, so kinetic energy is zero; its height and potential energy are greatest. Gravity still acts downward, so force and acceleration are not zero.
Answer
(iii) and (iv) are correct. Its speed is momentarily zero, so kinetic energy is zero; its height and potential energy are greatest. Gravity still acts downward, so force and acceleration are not zero.
Question Revise 4
Identify transformations: uphill truck; unwinding watch spring; photosynthesis; dam water; burning match; firecracker; microphone; bulb; solar panel.
Steps only
(i) Fuel chemical → kinetic and gravitational potential. (ii) Elastic potential → kinetic. (iii) Light → chemical. (iv) Gravitational potential → kinetic. (v) Chemical → heat and light. (vi) Chemical → heat, light, sound, and kinetic. (vii) Sound → electrical. (viii) Electrical → light and heat. (ix) Light → electrical.
Answer
(i) Fuel chemical → kinetic and gravitational potential.
(ii) Elastic potential → kinetic.
(iii) Light → chemical.
(iv) Gravitational potential → kinetic.
(v) Chemical → heat and light.
(vi) Chemical → heat, light, sound, and kinetic.
(vii) Sound → electrical.
(viii) Electrical → light and heat.
(ix) Light → electrical.
Question Revise 5
A 50 kg student reaches a height of 72.5 m by elevator, then by stairs. Find the potential-energy gain in both cases (g = 10 m s⁻²). Does the path matter?
Steps only
For either path, ΔU = mgh = 50 × 10 × 72.5 = 36,250 J. The gain is the same because only the vertical height change matters, not the path.
Answer
For either path, ΔU = mgh = 50 × 10 × 72.5 = 36,250 J. The gain is the same because only the vertical height change matters, not the path.
Question Revise 6
A crane raises the same mass to the 20th floor instead of the 10th, taking twice as long. How do energy and power compare?
Steps only
Assuming equal floor heights from ground, height doubles, so gravitational energy mgh doubles. Time also doubles, so average power (energy/time) stays the same.
Answer
Assuming equal floor heights from ground, height doubles, so gravitational energy mgh doubles. Time also doubles, so average power (energy/time) stays the same.
Question Revise 7
What determines the energy needed to raise a flag? Does speed affect work? What if speed doubles?
Steps only
Work against gravity depends on flag mass and height: W = mgh (ignoring friction). Raising it faster does not change this work, but it takes less time. Since power = work/time, doubling lifting speed (same path in half the time) doubles power.
Answer
Work against gravity depends on flag mass and height: W = mgh (ignoring friction). Raising it faster does not change this work, but it takes less time. Since power = work/time, doubling lifting speed (same path in half the time) doubles power.
Question Revise 8
A 60 kg man rides a 100 kg scooter to speed v; next day a 40 kg son joins. Same speed and time. What ratio of fuel energy is used?
Steps only
First total mass = 160 kg; with son = 200 kg. For the same speed, kinetic energy is proportional to total mass. Energy (and ideal fuel use) ratio first day:second day = 160:200 = 4:5.
Answer
First total mass = 160 kg; with son = 200 kg. For the same speed, kinetic energy is proportional to total mass. Energy (and ideal fuel use) ratio first day:second day = 160:200 = 4:5.
Question Revise 9
An adult weighs twice as much as a child. How far should each sit from the fulcrum for balance?
Steps only
Balance requires weight × distance equal on both sides. The adult must sit half as far from the fulcrum as the child. Example: child 2 m away, adult 1 m away.
Answer
Balance requires weight × distance equal on both sides. The adult must sit half as far from the fulcrum as the child. Example: child 2 m away, adult 1 m away.
Question Revise 10
A 2 kg ball is thrown upward at 20 m s⁻¹ and reaches 19.4 m. (i) Sign of gravity’s work on ascent/descent? (ii) Work by air resistance? Take g = 10.
Steps only
(i) Gravity does negative work on the way up and positive work on the way down. (ii) Initial kinetic energy = ½ × 2 × 20² = 400 J. Gain in gravitational potential energy = mgh = 2 × 10 × 19.4 = 388 J. At the top, K ≈ 0, so total change in mechanical energy is 388 − 400 = −12 J. Air resistance did −12 J of work.
Answer
(i) Gravity does negative work on the way up and positive work on the way down.
(ii) Initial kinetic energy = ½ × 2 × 20² = 400 J. Gain in gravitational potential energy = mgh = 2 × 10 × 19.4 = 388 J. At the top, K ≈ 0, so total change in mechanical energy is 388 − 400 = −12 J. Air resistance did −12 J of work.
Question Revise 11
A 10 kg block has initial kinetic energy 180 J. Fig. 7.37 gives the applied force from 0 to 4 m. Find speed at 0 m and 4 m; when is acceleration negative?
Steps only
Work is the area under the graph: 25 J + 100 J + 25 J = 150 J. Final kinetic energy is 330 J, so the speed at 4 m is √66 ≈ 8.12 m s⁻¹.
Answer
Work is the area under the graph: 25 J + 100 J + 25 J = 150 J. Final kinetic energy is 330 J, so the speed at 4 m is √66 ≈ 8.12 m s⁻¹.
Question Revise 12
A ball reaches 8 m on Earth. How high will it rise on the Moon where gravity is 1/6 as strong, with the same launch speed?
Steps only
For a vertical throw, maximum height h = v²/(2g). With the same initial speed and gMoon = gEarth/6, height is six times larger: 6 × 8 = 48 m.
Answer
For a vertical throw, maximum height h = v²/(2g). With the same initial speed and gMoon = gEarth/6, height is six times larger: 6 × 8 = 48 m.
Question Revise 13
A 1000 kg car travels at constant speed then brakes as shown in Fig. 7.38. Describe A–B, find kinetic energy at A, work by brakes B–C, and energy transformation.
Steps only
A–B is the reaction interval: speed stays at 35 m s⁻¹. Kinetic energy at A = ½ × 1000 × 35² = 612,500 J. During braking B–C, the car stops, so work by brakes equals change in kinetic energy = −612,500 J. The lost kinetic energy mainly becomes heat and sound.
Answer
A–B is the reaction interval: speed stays at 35 m s⁻¹. Kinetic energy at A = ½ × 1000 × 35² = 612,500 J. During braking B–C, the car stops, so work by brakes equals change in kinetic energy = −612,500 J. The lost kinetic energy mainly becomes heat and sound.
Question Revise 14
A 0.5 kg ball moves frictionlessly; at O, speed is zero and PE = 30 J. Use Fig. 7.39 to find speed at P, Q, R.
Steps only
P: 6.3 m s⁻¹; Q: 0 m s⁻¹; R: cannot be reached (40 J needed, total energy is 30 J).
Answer
P: 6.3 m s⁻¹; Q: 0 m s⁻¹; R cannot be reached.
Question Revise 15
A 1.5 kg coconut falls 10 m and stops in wet sand. Find its impact speed and depression depth if sand’s average resistance is 3000 N; g=10.
Steps only
Before impact, mgh becomes kinetic energy: v = √(2gh) = √200 ≈ 14.1 m s⁻¹. Energy at impact = 1.5 × 10 × 10 = 150 J. The sand’s work stops it: 3000 × depth = 150, so depth = 0.05 m = 5 cm.
Answer
Before impact, mgh becomes kinetic energy: v = √(2gh) = √200 ≈ 14.1 m s⁻¹. Energy at impact = 1.5 × 10 × 10 = 150 J. The sand’s work stops it: 3000 × depth = 150, so depth = 0.05 m = 5 cm.
Question The Journey Beyond 1
In the rubber-band pen model, what happens when you stretch and release the band? How does stretch affect distance?
Steps only
Stretching stores elastic potential energy. On release it changes into kinetic energy of the refill. A greater stretch usually stores more energy, so the refill travels farther, provided the setup and friction stay similar. Use eye protection and a safe direction.
Answer
Stretching stores elastic potential energy. On release it changes into kinetic energy of the refill. A greater stretch usually stores more energy, so the refill travels farther, provided the setup and friction stay similar. Use eye protection and a safe direction.
Question The Journey Beyond 2
How can you test a homemade lever, pulley, or ramp?
Steps only
Use a small safe load. Measure the load and the effort needed to move it. Mechanical advantage = load/effort. Compare different designs and note that a smaller effort usually requires movement over a longer distance.
Answer
Use a small safe load. Measure the load and the effort needed to move it. Mechanical advantage = load/effort. Compare different designs and note that a smaller effort usually requires movement over a longer distance.
Question The Journey Beyond 3
What can an energy simulation help you observe?
Steps only
Change mass, starting height, or friction and track how potential, kinetic, and thermal energy change. With little friction, potential and kinetic energy trade while total mechanical energy stays nearly constant. With friction, some becomes thermal energy.
Answer
Change mass, starting height, or friction and track how potential, kinetic, and thermal energy change. With little friction, potential and kinetic energy trade while total mechanical energy stays nearly constant. With friction, some becomes thermal energy.