Science questions, chapter 7

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Energy and power

44 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. Take g = 10 m s⁻² (weight of 1 kg = 10 N).

Section A: objective questions

12 marks

Choose the best answer for each question.

Question 1

Which statement defines work in science?

The ability to do work
The product of force and displacement in the direction of the force
The rate at which energy is transferred
The product of mass and gravitational acceleration

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Work is force × displacement in the direction of the force, W = Fs. The ability to do work is energy; the rate of doing work is power; mass × g is weight.

For the teacher: Remembering, low difficulty, standard 7.1.1

Textbook p. 210

Question 2

Which is the S.I. unit of power?

Joule (J)
Newton (N)
Joule second (J s)
Watt (W)

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Power is measured in watts. 1 W = 1 J s⁻¹, that is 1 J of work done in 1 s.

For the teacher: Remembering, low difficulty, standard 7.1.2

Textbook p. 212

Question 3

A student pushes a trolley with a force of 50 N over a distance of 4 m in the direction of the force. How much work is done?

12.5 J
54 J
200 J
46 J

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W = Fs = 50 N × 4 m = 200 J.

For the teacher: Applying, low difficulty, standard 7.1.1

Textbook p. 211

Question 4

A machine does 600 J of work in 20 s. What is its power?

30 W
620 W
12 000 W
580 W

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P = W ÷ t = 600 J ÷ 20 s = 30 W.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 213

Question 5

In which situation is work done in the scientific sense?

A student holds a heavy bag still for one minute
A boy pulls a box across the floor over a distance of 5 m
A girl pushes hard against a wall that does not move
A book rests on a table

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Work is done only when the object is displaced in the direction of the force. Holding a bag still, pushing a fixed wall and a resting book all have zero displacement.

For the teacher: Understanding, low difficulty, standard 7.1.1

Textbook p. 210

Question 6

A stone of mass 2 kg is lifted to a height of 5 m above the ground. What is its gravitational potential energy?

10 J
20 J
100 J
200 J

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E = mgh = 2 kg × 10 m s⁻² × 5 m = 100 J.

For the teacher: Applying, low difficulty, standard 7.2.1

Textbook p. 216

Question 7

A trolley of mass 4 kg moves at 3 m s⁻¹. What is its kinetic energy?

6 J
12 J
36 J
18 J

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E = ½mv² = ½ × 4 × 3² = ½ × 4 × 9 = 18 J. (Not 12 J: the speed must be squared first.)

For the teacher: Applying, low difficulty, standard 7.2.3

Textbook p. 220

Question 8

A spring is compressed by 5 cm by a final force of 40 N. What is the elastic potential energy stored?

1 J
2 J
100 J
200 J

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x = 5 cm = 0.05 m. E = ½Fx = ½ × 40 × 0.05 = 1 J.

For the teacher: Applying, low difficulty, standard 7.2.2

Textbook p. 218

Question 9

A simple pendulum bob swings in a closed system. What is true when the bob is at the lowest point?

Gravitational P.E. is maximum and K.E. is maximum
Gravitational P.E. is maximum and K.E. is zero
Gravitational P.E. and K.E. are both zero
Gravitational P.E. is minimum and K.E. is maximum

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At the lowest point the height is least, so P.E. is minimum, and the speed is greatest, so K.E. is maximum.

For the teacher: Understanding, low difficulty, standard 7.3.1

Textbook p. 223

Question 10

A car of mass 1 000 kg travels at 72 km h⁻¹. What is its kinetic energy?

36 000 J
200 000 J
720 000 J
2 592 000 J

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72 km h⁻¹ = 72 ÷ 3.6 = 20 m s⁻¹. E = ½ × 1 000 × 20² = 200 000 J.

For the teacher: Applying, medium difficulty, standard 7.2.3

Textbook p. 220

Question 11

A ball moves at speed v and has kinetic energy E. The speed of the ball is then doubled. What is its new kinetic energy?

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E = ½mv². If v becomes 2v, then (2v)² = 4v², so the kinetic energy becomes 4E.

For the teacher: Analysing, medium difficulty, standard 7.2.3

Textbook p. 220

Question 12

A real pendulum swings a little lower each time and finally stops. Which explanation obeys the Principle of Conservation of Energy?

The energy is changed into heat by friction and air resistance
The energy is destroyed as the pendulum swings
The energy is changed into more potential energy
The energy of the pendulum is created again by gravity

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Energy cannot be destroyed. In a real system some energy changes into heat (wasted energy) through friction and air resistance. A closed system has no such loss.

For the teacher: Understanding, low difficulty, standard 7.3.2

Textbook p. 225

Section B: structured questions

32 marks

Answer all questions. Show your working and include units.

Question 13

Write the formula for each form of energy. m is mass, g is gravitational acceleration, h is height, F is force, x is displacement and v is velocity.

[1 mark] (a)Gravitational potential energy

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mgh1

Tick each point you got.

[1 mark] (b)Elastic potential energy

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½Fx1

Tick each point you got.

[1 mark] (c)Kinetic energy

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½mv²1

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 7.2.1

Textbook p. 216

Question 14

An electric motor lifts a load of weight 3 000 N through a vertical height of 5 m in 40 s.

[2 marks] (a)Calculate the work done by the motor.

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W = Fs = 3 000 × 5 (correct substitution)1
= 15 000 J (answer with unit)1

Tick each point you got.

[1 mark] (b)How much energy is used by the motor to lift the load?

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15 000 J (energy used = work done)1

Tick each point you got.

[2 marks] (c)Calculate the power of the motor.

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P = W ÷ t = 15 000 ÷ 40 (correct substitution)1
= 375 W (answer with unit)1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 213

Question 15

Diagram 1 shows a graph of the force against the compression of a spring.

Force-compression graph of a spring: a straight line from the origin to 12 N at 0.10 m — Diagram 1

[2 marks] (a)Calculate the elastic potential energy stored in the spring when it is compressed by 0.10 m.

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E = ½Fx = ½ × 12 × 0.10 (or area under the graph)1
= 0.6 J (answer with unit)1

Tick each point you got.

[2 marks] (b)The spring is released and pushes a ball of mass 0.3 kg. Calculate the maximum speed of the ball.

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½mv² = 0.6, so ½ × 0.3 × v² = 0.6 and v² = 41
v = 2 m s⁻¹ (answer with unit)1

Tick each point you got.

[1 mark] (c)State one assumption made in part (b).

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No energy is lost to the surroundings (all the elastic P.E. becomes kinetic energy of the ball)1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 7.2.2

Textbook p. 219

Question 16

Diagram 2 shows a pendulum bob of mass 100 g released from A, 20 cm above its lowest point B. Take the oscillation as a closed system.

A pendulum bob released at A, 20 cm above the lowest point B — Diagram 2

[2 marks] (a)Calculate the gravitational potential energy of the bob at A relative to B.

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E = mgh = 0.1 × 10 × 0.20 (mass 100 g = 0.1 kg, 20 cm = 0.20 m)1
= 0.2 J (answer with unit)1

Tick each point you got.

[1 mark] (b)State the form of energy that is greatest at B.

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Kinetic energy1

Tick each point you got.

[2 marks] (c)Calculate the maximum speed of the bob.

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Kinetic energy at B = P.E. at A: ½ × 0.1 × v² = 0.2, so v² = 41
v = 2 m s⁻¹ (answer with unit)1

Tick each point you got.

[2 marks] (d)In a real pendulum the bob does not return to the height of A after many swings. Explain why, using the Principle of Conservation of Energy.

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Some energy is changed into heat (by friction at the pivot and air resistance)1
Energy is not destroyed; the total energy stays the same but less is available as P.E. and K.E.1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 7.3.2

Textbook p. 225

Question 17

Aina weighs 450 N and Bala weighs 600 N. Each climbs the same flight of stairs of vertical height 4 m. Aina takes 12 s and Bala takes 20 s.

[2 marks] (a)Calculate the work done by each student.

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Aina: 450 × 4 = 1 800 J1
Bala: 600 × 4 = 2 400 J1

Tick each point you got.

[2 marks] (b)Calculate the power of each student.

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Aina: 1 800 ÷ 12 = 150 W1
Bala: 2 400 ÷ 20 = 120 W1

Tick each point you got.

[1 mark] (c)Bala does more work. Explain why Aina still has the higher power.

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Power is the rate of doing work; Aina did her work in a shorter time (150 W is more than 120 W)1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 7.1.2

Textbook p. 213

Question 18

A load hangs on a spring and oscillates up and down. X is the position where the spring is most stretched and Y is the equilibrium position.

[1 mark] (a)At X, the elastic potential energy of the system is

AMaximumBMinimumCZero

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Maximum

The spring is most stretched, so elastic P.E. is maximum.

[1 mark] (b)At X, the kinetic energy of the load is

AMaximumBZeroCHalf of its maximum

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Zero

The load is momentarily at rest at the end of its motion.

[1 mark] (c)At Y, the kinetic energy of the load is

AMaximumBMinimumCZero

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Maximum

At the equilibrium position the load moves fastest.

For the teacher: Understanding, low difficulty, standard 7.3.1

Textbook p. 224

Question 19

You are asked to build a working model of a roller coaster using a rubber hose, a ball bearing and retort stands.

[1 mark] (a)State where the starting point of the ball bearing should be placed, and why.

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At the highest point of the track, so the ball has the most gravitational potential energy to change into kinetic energy1

Tick each point you got.

[2 marks] (b)Describe the energy changes of the ball bearing as it rolls down the first slope and then up a second hill.

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Down the slope: gravitational P.E. decreases and K.E. increases1
Up the hill: K.E. decreases and gravitational P.E. increases1

Tick each point you got.

[1 mark] (c)Explain why, in the real model, the second hill must be lower than the starting point.

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Some energy is changed to heat by friction, so the ball does not have enough energy to reach a height equal to or above the start1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 7.3.2

Textbook p. 229

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