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A Level H1 Physics Energy Power Quiz

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A-Level Physics H1 Quiz - Energy Power

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 60

Duration: 45 minutes
Total Marks: 60

Instructions:

Answer all questions.
Show all working clearly. Marks are awarded for correct methods even if the final answer is incorrect.
Use $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.
State units in all final answers.

Section A: Multiple Choice & Short Concepts (Questions 1–5)

[1 mark each]

1. Which of the following defines the watt (W) in terms of SI base units? A. $\text{kg m s}^{-2}$ B. $\text{kg m}^2 \text{ s}^{-3}$ C. $\text{kg m}^2 \text{ s}^{-2}$ D. $\text{J s}^{-1}$

2. A force of $10 \text{ N}$ acts on a body, moving it $5 \text{ m}$ in the direction of the force. What is the work done? A. $2 \text{ J}$ B. $50 \text{ J}$ C. $15 \text{ J}$ D. $0.5 \text{ J}$

3. A car travels at a constant velocity of $20 \text{ m s}^{-1}$ against a constant resistive force of $500 \text{ N}$ . What is the power output of the engine? A. $25 \text{ W}$ B. $500 \text{ W}$ C. $10,000 \text{ W}$ D. $20,000 \text{ W}$

4. Which statement about efficiency is correct? A. Efficiency can exceed $100\%$ if the machine is well-lubricated. B. Efficiency is the ratio of useful power output to total power input. C. Efficiency is the ratio of total energy input to useful energy output. D. Efficiency has units of Watts.

5. A ball is dropped from a height. As it falls, air resistance increases. Which graph best represents the variation of kinetic energy with time? A. A straight line through the origin. B. A curve with increasing gradient. C. A curve with decreasing gradient, approaching a constant value. D. A horizontal line.

Section B: Structured Calculations (Questions 6–15)

[Marks vary as indicated]

6. A crane lifts a load of mass $200 \text{ kg}$ vertically through a height of $15 \text{ m}$ in $12 \text{ s}$ . (a) Calculate the work done against gravity. [2]

(b) Calculate the average power output of the crane motor. [2]

7. A block of mass $4.0 \text{ kg}$ is pulled along a rough horizontal surface by a horizontal force of $20 \text{ N}$ . The block moves at a constant velocity of $3.0 \text{ m s}^{-1}$ . (a) State the magnitude of the frictional force acting on the block. [1]

(b) Calculate the power dissipated by friction. [2]

8. A student pushes a box of mass $10 \text{ kg}$ up a smooth inclined plane. The plane is inclined at $30^\circ$ to the horizontal. The box moves a distance of $5.0 \text{ m}$ along the slope. (a) Calculate the gain in gravitational potential energy of the box. [3]

(b) If the student pushes the box with a force parallel to the slope, calculate the minimum work done by the student. [2]

9. An electric motor lifts a $50 \text{ kg}$ mass vertically at a constant speed of $2.0 \text{ m s}^{-1}$ . The motor is connected to a $240 \text{ V}$ supply and draws a current of $5.0 \text{ A}$ . (a) Calculate the useful power output of the motor. [2]

(b) Calculate the efficiency of the motor. [2]

10. A car of mass $1200 \text{ kg}$ accelerates from rest to $25 \text{ m s}^{-1}$ in $10 \text{ s}$ . Assume air resistance is negligible during this acceleration. (a) Calculate the gain in kinetic energy of the car. [2]

(b) Determine the average power developed by the engine during this interval. [2]

11. A hydroelectric power station uses water falling from a height of $80 \text{ m}$ to drive turbines. The flow rate of water is $200 \text{ kg s}^{-1}$ . (a) Calculate the theoretical maximum power available from the falling water. [3]

(b) The actual electrical power output is $140 \text{ kW}$ . Calculate the efficiency of the power station. [2]

12. A spring obeys Hooke’s Law. A force of $10 \text{ N}$ extends the spring by $4.0 \text{ cm}$ . (a) Calculate the spring constant $k$ . [2]

(b) Calculate the elastic potential energy stored in the spring when extended by $4.0 \text{ cm}$ . [2]

13. A pendulum bob of mass $0.5 \text{ kg}$ is released from rest at a height of $0.2 \text{ m}$ above its lowest point. (a) Calculate the speed of the bob at the lowest point, assuming no air resistance. [3]

(b) In reality, the bob only reaches a height of $0.18 \text{ m}$ on the other side. Calculate the energy lost to air resistance and friction in one swing. [2]

14. A cyclist travels at a constant speed of $8.0 \text{ m s}^{-1}$ on a level road. The total resistive force (air resistance + friction) is $40 \text{ N}$ . (a) Calculate the power output of the cyclist. [2]

(b) The cyclist stops pedaling. Explain, in terms of energy transformations, why the bicycle eventually stops. [2]

15. A pump raises $500 \text{ kg}$ of water from a well $10 \text{ m}$ deep in $20 \text{ s}$ . (a) Calculate the work done by the pump. [2]

(b) If the pump is $80\%$ efficient, calculate the input power required. [3]

Section C: Data Analysis & Reasoning (Questions 16–20)

[Marks vary as indicated]

16. The graph below shows the variation of force $F$ with extension $x$ for a rubber band. (Imagine a graph where the loading curve is non-linear and the unloading curve is below the loading curve, forming a hysteresis loop.)

(a) Explain what the area under the loading curve represents. [1]

(b) Explain the significance of the area enclosed between the loading and unloading curves. [2]

17. A car engine has a maximum power output of $100 \text{ kW}$ . The car has a mass of $1500 \text{ kg}$ . (a) Explain why the acceleration of the car decreases as its speed increases, even if the engine is delivering maximum power. [3]

(b) Calculate the maximum theoretical speed the car could reach if the total resistive force at that speed is $2000 \text{ N}$ . [2]

18. Consider a system where a ball bounces on the floor. (a) Define an elastic collision. [1]

(b) A ball is dropped from $1.0 \text{ m}$ and rebounds to $0.8 \text{ m}$ . Is this collision elastic? Justify your answer with a calculation of the coefficient of restitution or energy ratio. [3]

19. An electric heater is rated at $2.0 \text{ kW}$ . It is used to heat $2.0 \text{ kg}$ of water. The specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$ . (a) Calculate the time taken to raise the temperature of the water by $50^\circ\text{C}$ , assuming no energy loss. [3]

(b) In practice, the time taken is longer. State one reason for this discrepancy. [1]

20. A roller coaster car of mass $500 \text{ kg}$ starts from rest at the top of a hill of height $40 \text{ m}$ . It travels down the track and up a second hill of height $25 \text{ m}$ . (a) Calculate the speed of the car at the top of the second hill, assuming the track is frictionless. [4]

(b) If the actual speed at the top of the second hill is $15 \text{ m s}^{-1}$ , calculate the work done against friction and air resistance during the journey. [3]

Answers

A-Level Physics H1 Quiz - Energy Power (Answer Key)

1. B

Power $P = \frac{W}{t} = \frac{F d}{t}$ .
Units: $\frac{(\text{kg m s}^{-2})(\text{m})}{\text{s}} = \text{kg m}^2 \text{ s}^{-3}$ .

2. B

$W = F d = 10 \times 5 = 50 \text{ J}$ .

3. C

$P = F v = 500 \times 20 = 10,000 \text{ W}$ .

4. B

Efficiency = $\frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\%$ . It is a ratio and has no units. It cannot exceed $100\%$ .

5. C

Initially, KE increases as speed increases. As air resistance increases, acceleration decreases, so the rate of gain of KE decreases. Eventually, terminal velocity is reached, and KE becomes constant.

6. (a) Work done against gravity = Gain in GPE $W = mgh = 200 \times 9.81 \times 15$ $W = 29,430 \text{ J} \quad (\text{or } 29.4 \text{ kJ})$ [1 for formula, 1 for answer]

(b) Average Power $P = \frac{W}{t} = \frac{29,430}{12}$ $P = 2,452.5 \text{ W} \quad (\text{or } 2.45 \text{ kW})$ [1 for formula, 1 for answer]

7. (a) Since velocity is constant, acceleration is zero. By Newton's First Law, the net force is zero. Therefore, Frictional Force = Applied Force = $20 \text{ N}$ . [1]

(b) Power dissipated by friction $P = F v = 20 \times 3.0$ $P = 60 \text{ W}$ [1 for formula, 1 for answer]

8. (a) Vertical height gained $h = d \sin \theta = 5.0 \sin 30^\circ = 2.5 \text{ m}$ . $\Delta \text{GPE} = mgh = 10 \times 9.81 \times 2.5$ $\Delta \text{GPE} = 245.25 \text{ J} \quad (\text{accept } 245 \text{ J})$ [1 for height, 1 for formula, 1 for answer]

(b) On a smooth slope, Work Done = Gain in GPE (Conservation of Energy). $W = 245 \text{ J}$ [1 for reasoning, 1 for answer]

9. (a) Useful Power Output (lifting power) $P_{\text{out}} = F v = mg v = 50 \times 9.81 \times 2.0$ $P_{\text{out}} = 981 \text{ W}$ [1 for formula, 1 for answer]

(b) Input Power $P_{\text{in}} = VI = 240 \times 5.0 = 1200 \text{ W}$ Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{981}{1200} \times 100\%$ $\eta = 81.75\% \quad (\text{accept } 82\%)$ [1 for input power, 1 for efficiency calc]

10. (a) Gain in Kinetic Energy $\Delta \text{KE} = \frac{1}{2} m v^2 - 0 = \frac{1}{2} (1200) (25)^2$ $\Delta \text{KE} = 600 \times 625 = 375,000 \text{ J} \quad (\text{or } 375 \text{ kJ})$ [1 for formula, 1 for answer]

(b) Average Power $P = \frac{\Delta E}{t} = \frac{375,000}{10}$ $P = 37,500 \text{ W} \quad (\text{or } 37.5 \text{ kW})$ [1 for formula, 1 for answer]

11. (a) Mass flow rate $\frac{dm}{dt} = 200 \text{ kg s}^{-1}$ . Power available = Rate of loss of GPE $P = \frac{mgh}{t} = \left(\frac{m}{t}\right) gh = 200 \times 9.81 \times 80$ $P = 156,960 \text{ W} \quad (\text{or } 157 \text{ kW})$ [1 for concept, 1 for substitution, 1 for answer]

(b) Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{140,000}{156,960}$ $\eta = 0.8919 \dots \approx 89.2\%$ [1 for formula, 1 for answer]

12. (a) Hooke's Law: $F = kx$ $10 = k (0.04)$ $k = \frac{10}{0.04} = 250 \text{ N m}^{-1}$ [1 for formula, 1 for answer]

(b) Elastic Potential Energy $E = \frac{1}{2} F x = \frac{1}{2} (10) (0.04)$ $E = 0.2 \text{ J}$ (Alternatively $E = \frac{1}{2} k x^2 = 0.5 \times 250 \times 0.04^2 = 0.2 \text{ J}$ ) [1 for formula, 1 for answer]

13. (a) Conservation of Energy: Loss in GPE = Gain in KE $mgh = \frac{1}{2} m v^2 \Rightarrow v = \sqrt{2gh}$ $v = \sqrt{2 \times 9.81 \times 0.2} = \sqrt{3.924}$ $v = 1.98 \text{ m s}^{-1}$ [1 for principle, 1 for substitution, 1 for answer]

(b) Initial Energy (relative to lowest point) = $mgh_1$ . Final Energy = $mgh_2$ . Energy Lost = $mg(h_1 - h_2)$ $E_{\text{lost}} = 0.5 \times 9.81 \times (0.2 - 0.18)$ $E_{\text{lost}} = 0.5 \times 9.81 \times 0.02 = 0.0981 \text{ J}$ [1 for concept, 1 for answer]

14. (a) Power output $P = F v = 40 \times 8.0 = 320 \text{ W}$ [1 for formula, 1 for answer]

(b) When pedaling stops, the driving force is removed. The kinetic energy of the bicycle is gradually converted into thermal energy (heat) and sound due to work done against resistive forces (friction and air resistance). When all KE is dissipated, the bicycle stops. [1 for KE conversion, 1 for resistive forces]

15. (a) Work done $W = mgh = 500 \times 9.81 \times 10 = 49,050 \text{ J}$ [1 for formula, 1 for answer]

(b) Useful Power Output $P_{\text{out}} = \frac{W}{t} = \frac{49,050}{20} = 2,452.5 \text{ W}$ Efficiency $\eta = 0.80 = \frac{P_{\text{out}}}{P_{\text{in}}}$ $P_{\text{in}} = \frac{P_{\text{out}}}{0.80} = \frac{2,452.5}{0.8}$ $P_{\text{in}} = 3,065.6 \text{ W} \quad (\text{or } 3.07 \text{ kW})$ [1 for useful power, 1 for efficiency rearrangement, 1 for answer]

16. (a) The area under the loading curve represents the work done to stretch the rubber band (or elastic potential energy stored). [1]

(b) The area enclosed by the loop (hysteresis loop) represents the energy dissipated as heat (thermal energy) during the loading and unloading cycle. This is why rubber bands get warm when stretched repeatedly. [1 for identification, 1 for explanation]

17. (a) Power $P = F v$ . If $P$ is constant (max power), then Driving Force $F = \frac{P}{v}$ . As speed $v$ increases, the driving force $F$ decreases. Since Resistive Force increases with speed (or is constant), the Net Force ( $F_{\text{net}} = F_{\text{drive}} - F_{\text{resist}}$ ) decreases. Since $a = \frac{F_{\text{net}}}{m}$ , acceleration decreases. [1 for $F=P/v$ , 1 for net force decrease, 1 for link to acceleration]

(b) At maximum speed, acceleration is zero, so Driving Force = Resistive Force. $F_{\text{drive}} = 2000 \text{ N}$ $P = F v \Rightarrow 100,000 = 2000 \times v$ $v = \frac{100,000}{2000} = 50 \text{ m s}^{-1}$ [1 for condition, 1 for answer]

18. (a) An elastic collision is one in which kinetic energy is conserved (total KE before = total KE after). [1]

(b) No, it is not elastic. $\text{KE}_{\text{initial}} \propto h_{\text{initial}} = 1.0 \text{ m}$ . $\text{KE}_{\text{final}} \propto h_{\text{final}} = 0.8 \text{ m}$ . Since $h_{\text{final}} < h_{\text{initial}}$ , KE is lost. Ratio of KE = $0.8/1.0 = 0.8$ . Since KE is not conserved ( $80\%$ remains), it is inelastic. [1 for conclusion, 1 for comparison/calc, 1 for justification]

19. (a) Energy required $Q = mc\Delta \theta$ $Q = 2.0 \times 4200 \times 50 = 420,000 \text{ J}$ Power $P = 2000 \text{ W}$ . $t = \frac{E}{P} = \frac{420,000}{2000} = 210 \text{ s}$ [1 for energy calc, 1 for formula, 1 for answer]

(b) Energy is lost to the surroundings (heating the container, air) or the heater itself. [1]

20. (a) Conservation of Energy between Top 1 and Top 2: $\text{GPE}_1 + \text{KE}_1 = \text{GPE}_2 + \text{KE}_2$ Take reference level at bottom ( $h=0$ ). $mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2$ Cancel $m$ : $gh_1 = gh_2 + \frac{1}{2}v^2$ $9.81(40) = 9.81(25) + 0.5 v^2$ $392.4 = 245.25 + 0.5 v^2$ $147.15 = 0.5 v^2$ $v^2 = 294.3$ $v = 17.15 \text{ m s}^{-1} \quad (\text{accept } 17.2 \text{ m s}^{-1})$ [1 for principle, 1 for substitution, 1 for algebra, 1 for answer]

(b) Actual $\text{KE}_2 = \frac{1}{2}(500)(15)^2 = 56,250 \text{ J}$ . Theoretical $\text{KE}_2$ (from part a) $= \frac{1}{2}(500)(17.15)^2 \approx 73,575 \text{ J}$ . (Or use Energy difference directly): Total Energy at Start $= mgh_1 = 500 \times 9.81 \times 40 = 196,200 \text{ J}$ . Total Energy at End $= mgh_2 + \text{KE}_{\text{actual}} = (500 \times 9.81 \times 25) + 56,250$ $= 122,625 + 56,250 = 178,875 \text{ J}$ . Work done against friction = Energy Lost $W_f = 196,200 - 178,875 = 17,325 \text{ J} \quad (\text{or } 17.3 \text{ kJ})$ [1 for initial energy, 1 for final energy, 1 for difference]

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