A Level H1 Physics Energy Power Quiz

A-Level Physics H1 Quiz - Energy Power

Answer Key

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Question 1 [2 marks]

Answer:
The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Marking:

• [B1] for stating that work done equals change in kinetic energy
• [B1] for correct mathematical expression or clear verbal statement linking net work to kinetic energy change

Teaching notes: This is a fundamental principle connecting the concept of work (a process quantity) to kinetic energy (a state quantity). It means that if positive net work is done on an object, it speeds up; if negative net work is done, it slows down.

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Question 2 [2 marks]

Answer:
Power is defined as the rate of doing work (or the rate of energy transfer).

$P = \frac{W}{t} = \frac{\Delta E}{t}$

The SI unit of power is the watt (W), where $1 \text{ W} = 1 \text{ J s}^{-1}$.

Marking:

• [B1] for "rate of doing work" or "work done per unit time"
• [B1] for correct formula $P = W/t$ or $P = \Delta E/t$

Teaching notes: Power measures how quickly energy is transferred or work is done. A more powerful machine does the same amount of work in less time.

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Question 3 [2 marks]

Answer:

$W = mgh = 2.0 \times 9.81 \times 5.0 = 98.1 \text{ J}$

Since the object moves at constant speed, the applied force equals the weight, and the work done against gravity is 98 J (to 2 s.f.).

Marking:

• [B1] for correct substitution into $W = mgh$
• [B1] for correct answer with unit (J)

Common mistake: Students sometimes confuse the applied force with net force. At constant speed, the net force is zero, but work is still done against gravity.

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Question 4 [2 marks]

Answer:
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy of an isolated system remains constant.

Marking:

• [B1] for "energy cannot be created or destroyed"
• [B1] for "only transferred/transformed from one form to another" OR "total energy of an isolated system is constant"

Teaching notes: This is one of the most fundamental laws in physics. In practice, energy often appears to be "lost" — but it has been converted to other forms (usually thermal energy) rather than destroyed.

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Question 5 [3 marks]

Answer:

• Elastic potential energy is the energy stored in an object when it is deformed (stretched or compressed) and can return to its original shape. Example: a stretched spring or a compressed rubber band.
• Gravitational potential energy is the energy stored in an object due to its position in a gravitational field (i.e., its height above a reference level). Example: a book on a shelf.

Marking:

• [B1] for correct definition of elastic potential energy
• [B1] for correct definition of gravitational potential energy
• [B1] for one valid example of each

Teaching notes: Both are forms of stored (potential) energy. The key difference is the cause: elastic PE arises from deformation of a material, while gravitational PE arises from position in a gravitational field.

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Question 6 [2 marks]

Answer:

$\text{Efficiency} = \frac{\text{useful output power}}{\text{input power}} \times 100\% = \frac{400}{500} \times 100\% = 80\%$

Marking:

• [B1] for correct formula or substitution
• [B1] for correct answer: 80%

Common mistake: Students sometimes divide input by output, giving an efficiency greater than 100%.

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Question 7 [2 marks]

Answer:
A non-renewable energy source is one that is consumed faster than it can be replenished naturally, and will eventually run out. Example: coal / natural gas / petroleum / nuclear fuel (uranium).

Marking:

• [B1] for correct definition (finite resource / cannot be replenished on a human timescale)
• [B1] for a valid example

Teaching notes: Non-renewable sources like fossil fuels took millions of years to form. Once used, they are effectively gone on any human timescale.

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Question 8 [4 marks]

(a) [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = \frac{1}{2} \times 1200 \times 400 = 2.4 \times 10^5 \text{ J}$

Answer: $2.4 \times 10^5 \text{ J}$ (or 240 kJ)

Marking:

• [B1] for correct substitution into $KE = \frac{1}{2}mv^2$
• [B1] for correct answer with unit

(b) [2 marks]

$P = \frac{W}{t} = \frac{\Delta KE}{t} = \frac{2.4 \times 10^5}{10} = 2.4 \times 10^4 \text{ W}$

Answer: $2.4 \times 10^4 \text{ W}$ (or 24 kW)

Marking:

• [B1] for using $P = \frac{\Delta KE}{t}$ or equivalent
• [B1] for correct answer with unit

Teaching notes: The work-energy theorem tells us the net work done equals the change in kinetic energy. Power is the rate at which this energy is transferred.

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Question 9 [2 marks]

Answer:
No machine can be 100% efficient because some input energy is always wasted, typically converted to thermal energy (heat) due to friction, air resistance, or electrical resistance. This wasted energy is dissipated to the surroundings.

Marking:

• [B1] for stating that some energy is always "wasted" or "lost"
• [B1] for identifying the reason (friction / resistance / heat dissipation)

Teaching notes: This connects to the second law of thermodynamics — in any real process, some energy becomes unavailable for useful work.

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Question 10 [3 marks]

Answer:

• As the ball rises, its kinetic energy decreases and is converted into gravitational potential energy. The ball slows down.
• At the maximum height, the ball's kinetic energy is zero (momentarily at rest) and gravitational potential energy is at its maximum.
• As the ball falls, gravitational potential energy is converted back into kinetic energy. The ball speeds up.
• At the original height (on return), the kinetic energy equals the initial kinetic energy (same speed as at launch, but in the opposite direction).

Marking:

• [B1] for describing KE → GPE during ascent
• [B1] for describing GPE → KE during descent
• [B1] for stating that total mechanical energy is conserved (or that speed at return equals launch speed)

Teaching notes: Since air resistance is ignored, mechanical energy (KE + GPE) is conserved throughout. This is a classic application of energy conservation.

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Question 11 [6 marks]

(a) [2 marks]

$GPE = mgh = 0.50 \times 9.81 \times 20 = 98.1 \text{ J}$

Answer: 98 J (to 2 s.f.)

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [3 marks]

By conservation of energy:

$GPE_{\text{top}} = KE_{\text{bottom}}$

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$

Answer: $19.8 \text{ m s}^{-1}$ (or $20 \text{ m s}^{-1}$ to 2 s.f.)

Marking:

• [B1] for stating conservation of energy / equating GPE to KE
• [B1] for correct substitution
• [B1] for correct answer with unit

(c) [1 mark]

Answer: Air resistance is negligible / no energy is lost to the surroundings / the stone falls freely under gravity.

Marking:

• [B1] for any valid assumption

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Question 12 [8 marks]

(a) [2 marks]

The horizontal component of the applied force:

$F_x = F \cos\theta = 60 \cos 30° = 60 \times 0.866 = 51.96 \text{ N}$

Work done by applied force:

$W = F_x \times d = 51.96 \times 8.0 = 415.7 \text{ J}$

Answer: 416 J (or 420 J to 2 s.f.)

Marking:

• [B1] for using $F\cos\theta$ to find horizontal component
• [B1] for correct answer with unit

(b) [1 mark]

$W_{\text{friction}} = f \times d = 25 \times 8.0 = 200 \text{ J}$

Answer: 200 J

Marking:

• [B1] for correct answer with unit

(c) [2 marks]

$W_{\text{net}} = W_{\text{applied}} - W_{\text{friction}} = 415.7 - 200 = 215.7 \text{ J}$

Answer: 216 J (or 220 J to 2 s.f.)

Marking:

• [B1] for subtracting friction work from applied work
• [B1] for correct answer

(d) [3 marks]

By the work-energy theorem:

$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - 0$

$215.7 = \frac{1}{2} \times 15 \times v^2$

$v^2 = \frac{2 \times 215.7}{15} = 28.76$

$v = \sqrt{28.76} = 5.36 \text{ m s}^{-1}$

Answer: $5.36 \text{ m s}^{-1}$ (or $5.4 \text{ m s}^{-1}$ to 2 s.f.)

Marking:

• [B1] for using work-energy theorem
• [B1] for correct substitution
• [B1] for correct answer with unit

Teaching notes: This question combines resolving forces, calculating work at an angle, and applying the work-energy theorem — a classic multi-step problem.

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Question 13 [6 marks]

(a) [2 marks]

Gravitational potential energy lost per second:

$E = \frac{mgh}{t} = 5.0 \times 10^3 \times 9.81 \times 80 = 3.924 \times 10^6 \text{ J s}^{-1}$

Answer: $3.92 \times 10^6 \text{ J s}^{-1}$ (or 3.92 MW)

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [3 marks]

$\text{Efficiency} = \frac{\text{useful output}}{\text{input}} \times 100\%$

$P_{\text{output}} = 0.70 \times 3.924 \times 10^6 = 2.747 \times 10^6 \text{ W}$

Answer: $2.75 \times 10^6 \text{ W}$ (or 2.75 MW)

Marking:

• [B1] for using efficiency formula
• [B1] for correct substitution
• [B1] for correct answer with unit

(c) [1 mark]

Answer: Energy is lost as heat due to friction in the turbines / sound energy / kinetic energy of water leaving the turbine / heat from electrical resistance in the generators.

Marking:

• [B1] for any valid reason

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Question 14 [6 marks]

(a) [2 marks]

$GPE = mgh = 60 \times 9.81 \times 12 = 7063.2 \text{ J}$

Answer: 7060 J (or 7.06 kJ)

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [2 marks]

$P = \frac{W}{t} = \frac{7063.2}{15} = 470.9 \text{ W}$

Answer: 471 W (or 470 W to 2 s.f.)

Marking:

• [B1] for using $P = W/t$
• [B1] for correct answer with unit

(c) [2 marks]

Answer: The student's body is not 100% efficient. Additional energy is used for body processes such as breathing, blood circulation, and maintaining body temperature. Energy is also lost as heat from the body. The muscles also do work to move the legs and arms, not just to gain height.

Marking:

• [B1] for stating that the body has additional energy requirements beyond the mechanical work of climbing
• [B1] for giving a specific example (heat loss / internal body processes / limb movement)

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Question 15 [6 marks]

(a) [1 mark]

At constant speed, tension equals weight:

$T = mg = 4.0 \times 9.81 = 39.24 \text{ N}$

Answer: 39.2 N

Marking:

• [B1] for correct answer with unit

(b) [2 marks]

$P_{\text{output}} = T \times v = 39.24 \times 0.50 = 19.62 \text{ W}$

Answer: 19.6 W

Marking:

• [B1] for using $P = Fv$
• [B1] for correct answer with unit

(c) [3 marks]

Input power:

$P_{\text{input}} = VI = 12 \times 1.2 = 14.4 \text{ W}$

Wait — this gives an efficiency greater than 100%, which is impossible. Let me recalculate:

$P_{\text{input}} = VI = 12 \times 1.2 = 14.4 \text{ W}$

$P_{\text{output}} = 19.62 \text{ W}$

This is inconsistent. The output power (19.62 W) exceeds the input power (14.4 W), which violates energy conservation. The question values need to be self-consistent. Let me re-examine: the output power is $Tv = mgv = 4.0 \times 9.81 \times 0.50 = 19.62$ W, and input is $12 \times 1.2 = 14.4$ W. This is physically impossible.

Revised approach for the answer key: The question as stated has inconsistent values. However, for the purpose of the answer key, we proceed with the calculation as the student would:

$\text{Efficiency} = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\% = \frac{19.62}{14.4} \times 100\%$

This exceeds 100%, indicating an error in the question data. For a valid question, the input power should exceed the output power.

Note to generator: The values in Q15 should be revised so that $P_{\text{input}} > P_{\text{output}}$. For example, if $I = 2.0$ A, then $P_{\text{input}} = 24$ W and efficiency $= 19.62/24 \times 100\% = 81.8\%$.

Corrected answer (assuming the question intends valid physics):

If we use the values as given and the student identifies the inconsistency:

$\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} = \frac{19.62}{14.4} \times 100\%$

Since this exceeds 100%, the data is inconsistent. A corrected version would use $I = 2.0$ A:

$P_{\text{input}} = 12 \times 2.0 = 24 \text{ W}$

$\text{Efficiency} = \frac{19.62}{24} \times 100\% = 81.8\%$

Marking (for corrected version with $I = 2.0$ A):

• [B1] for calculating $P_{\text{input}} = VI$
• [B1] for using efficiency formula
• [B1] for correct answer: 81.8% or 82%

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Question 16 [9 marks]

(a) [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.20 \times 6.0^2 = \frac{1}{2} \times 0.20 \times 36 = 3.6 \text{ J}$

Answer: 3.6 J

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [4 marks]

The height difference between the lowest and highest points is $2L = 2 \times 1.5 = 3.0$ m.

By conservation of energy:

$KE_{\text{lowest}} = KE_{\text{highest}} + GPE_{\text{highest}}$

$\frac{1}{2}mv_{\text{low}}^2 = \frac{1}{2}mv_{\text{high}}^2 + mg(2L)$

$\frac{1}{2}v_{\text{low}}^2 = \frac{1}{2}v_{\text{high}}^2 + g(2L)$

$\frac{1}{2}(36) = \frac{1}{2}v_{\text{high}}^2 + 9.81 \times 3.0$

$18 = \frac{1}{2}v_{\text{high}}^2 + 29.43$

$\frac{1}{2}v_{\text{high}}^2 = 18 - 29.43 = -11.43$

This gives a negative value, meaning the ball cannot reach the top of the circle with this speed. The minimum speed at the lowest point for the ball to just complete the circle is:

$v_{\text{min}} = \sqrt{5gL} = \sqrt{5 \times 9.81 \times 1.5} = \sqrt{73.575} = 8.58 \text{ m s}^{-1}$

Since $6.0 < 8.58$, the ball does not have enough speed to complete the vertical circle.

Answer: The ball does not have sufficient speed to reach the highest point of the circle. It will fall away from the circular path before reaching the top.

Marking:

• [B1] for applying conservation of energy with correct height difference ($2L$)
• [B1] for correct substitution
• [B1] for identifying that the result is physically impossible (negative $v^2$)
• [B1] for concluding that the ball cannot complete the circle

Teaching notes: This is a common exam question that tests whether students can recognise when a physical situation is impossible. The critical speed at the top of the circle is $\sqrt{gL}$, and using energy conservation, the minimum speed at the bottom is $\sqrt{5gL}$.

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Question 17 [10 marks]

(a) [2 marks]

$GPE_A = mgh_A = 400 \times 9.81 \times 30 = 1.177 \times 10^5 \text{ J}$

Answer: $1.18 \times 10^5 \text{ J}$ (or 118 kJ)

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [3 marks]

By conservation of energy between A and B:

$GPE_A = GPE_B + KE_B$

$mgh_A = mgh_B + \frac{1}{2}mv_B^2$

$v_B = \sqrt{2g(h_A - h_B)} = \sqrt{2 \times 9.81 \times (30 - 10)} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4}$

$v_B = 19.8 \text{ m s}^{-1}$

Answer: $19.8 \text{ m s}^{-1}$ (or $20 \text{ m s}^{-1}$ to 2 s.f.)

Marking:

• [B1] for applying conservation of energy
• [B1] for correct substitution
• [B1] for correct answer with unit

(c) [3 marks]

By conservation of energy between A and C (ground level, $h = 0$):

$GPE_A = KE_C$

$mgh_A = \frac{1}{2}mv_C^2$

$v_C = \sqrt{2gh_A} = \sqrt{2 \times 9.81 \times 30} = \sqrt{588.6} = 24.26 \text{ m s}^{-1}$

Answer: $24.3 \text{ m s}^{-1}$ (or $24 \text{ m s}^{-1}$ to 2 s.f.)

Marking:

• [B1] for applying conservation of energy
• [B1] for correct substitution
• [B1] for correct answer with unit

(d) [2 marks]

Answer: In practice, friction acts between the car and the track, and air resistance also opposes the motion. Some of the mechanical energy is converted to thermal energy (heat) and sound energy, so the kinetic energy at point C is less than the gravitational potential energy at point A.

Marking:

• [B1] for identifying friction and/or air resistance as the cause of energy loss
• [B1] for stating that energy is converted to heat/sound (thermal energy)

Teaching notes: This question demonstrates that while energy is always conserved, mechanical energy (KE + GPE) is only conserved when no non-conservative forces (like friction) do work.

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Question 18 [9 marks]

(a) [2 marks]

$GPE_{\text{max}} = mgh = 0.10 \times 9.81 \times 0.05 = 0.04905 \text{ J}$

Answer: $0.049 \text{ J}$ (or $4.9 \times 10^{-2}$ J)

Marking:

• [B1] for correct substitution
• [B1] for correct answer with unit

(b) [2 marks]

Answer: The maximum kinetic energy is 0.049 J. By conservation of energy, at the lowest point all the gravitational potential energy has been converted to kinetic energy, so $KE_{\text{max}} = GPE_{\text{max}}$.

Marking:

• [B1] for correct value
• [B1] for explaining using conservation of energy

(c) [3 marks]

$KE_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2$

$0.04905 = \frac{1}{2} \times 0.10 \times v_{\text{max}}^2$

$v_{\text{max}}^2 = \frac{2 \times 0.04905}{0.10} = 0.981$

$v_{\text{max}} = \sqrt{0.981} = 0.99 \text{ m s}^{-1}$

Answer: $0.99 \text{ m s}^{-1}$

Marking:

• [B1] for equating KE to GPE
• [B1] for correct substitution
• [B1] for correct answer with unit

(d) [2 marks]

Answer: The maximum speed remains the same. From $v = \sqrt{2gh}$, the maximum speed depends only on the height $h$ and $g$, not on the mass. Since the height is unchanged, the maximum speed is the same.

Marking:

• [B1] for stating that maximum speed is unchanged
• [B1] for explaining that $v = \sqrt{2gh}$ is independent of mass

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Question 19 [8 marks]

(a) [4 marks]

Device A: $\eta = \frac{150}{200} \times 100\% = 75\%$

Device B: $\eta = \frac{100}{500} \times 100\% = 20\%$

Device C: $\eta = \frac{640}{800} \times 100\% = 80\%$

Device D: $\eta = \frac{900}{1000} \times 100\% = 90\%$

Marking:

• [B1] for each correct efficiency (4 × [B1])

(b) [1 mark]

Answer: Device D is the most efficient (90%).

Marking:

• [B1] for correct answer

(c) [2 marks]

Wasted energy per second for each device:

• A: $200 - 150 = 50$ J
• B: $500 - 100 = 400$ J
• C: $800 - 640 = 160$ J
• D: $1000 - 900 = 100$ J

Total wasted: $50 + 400 + 160 + 100 = 710$ J

Answer: 710 J per second

Marking:

• [B1] for calculating individual wasted energies
• [B1] for correct total

(d) [1 mark]

Answer: The wasted energy is converted mainly to thermal energy (heat) due to friction in moving parts and/or electrical resistance in circuits. Some may also be converted to sound energy.

Marking:

• [B1] for identifying thermal energy / heat as the main waste form

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Question 20 [7 marks]

(a) [2 marks]

Volume of air passing through per second:

$V = A \times v \times t = 1.2 \times 10^3 \times 10 \times 1 = 1.2 \times 10^4 \text{ m}^3 \text{ (per second)}$

Mass of air per second:

$\frac{m}{t} = \rho \times A \times v = 1.2 \times 1.2 \times 10^3 \times 10 = 1.44 \times 10^4 \text{ kg s}^{-1}$

Answer: $1.44 \times 10^4 \text{ kg s}^{-1}$ ✓

Marking:

• [B1] for using mass flow rate = $\rho Av$
• [B1] for correct answer

(b) [2 marks]

$KE \text{ per second} = \frac{1}{2} \times \frac{m}{t} \times v^2 = \frac{1}{2} \times 1.44 \times 10^4 \times 10^2 = \frac{1}{2} \times 1.44 \times 10^4 \times 100 = 7.2 \times 10^5 \text{ J s}^{-1}$

Answer: $7.2 \times 10^5 \text{ W}$ (or 720 kW)

Marking:

• [B1] for using $KE = \frac{1}{2}mv^2$ with mass per second
• [B1] for correct answer with unit

(c) [2 marks]

$P_{\text{electrical}} = 0.35 \times 7.2 \times 10^5 = 2.52 \times 10^5 \text{ W}$

Answer: $2.52 \times 10^5 \text{ W}$ (or 252 kW)

Marking:

• [B1] for multiplying by efficiency
• [B1] for correct answer with unit

(d) [1 mark]

Answer: Density of air / cross-sectional area of the blades / number or shape of the blades / temperature of the air.

Marking:

• [B1] for any valid factor

(e) [2 marks]

Answer: If 100% of the wind's kinetic energy were extracted, the air would stop moving after passing through the turbine. This would cause air to pile up behind the turbine, preventing further wind from flowing through it. Therefore, it is physically impossible to extract all the kinetic energy — some must remain in the air to allow it to continue flowing. (This is described by Betz's law, which sets a theoretical maximum efficiency of 59.3% for wind turbines.)

Marking:

• [B1] for explaining that extracting all KE would stop the airflow
• [B1] for explaining the consequence (air cannot flow through the turbine / Betz limit concept)

Teaching notes: This is a conceptual question that goes beyond simple calculation. It tests whether students understand the physical limitations of energy extraction from a moving fluid.

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Total: 50 marks