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A Level H1 Physics Energy Power Quiz

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Questions

A-Level Physics H1 Quiz - Energy Power

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Show all working clearly. Marks are awarded for the method as well as the final answer.
Use $g = 9.81 \, \text{m s}^{-2}$ where appropriate.
The use of a scientific calculator is permitted.

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

Marks: 1 mark each

1. Which of the following is a correct unit for power 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. $\text{N m s}^{-1}$

2. A crane lifts a load of mass $500 \, \text{kg}$ vertically at a constant speed of $2.0 \, \text{m s}^{-1}$ . What is the useful power output of the crane? A. $1000 \, \text{W}$ B. $5000 \, \text{W}$ C. $9810 \, \text{W}$ D. $19620 \, \text{W}$

3. A ball is dropped from rest. Air resistance is negligible. Which graph best represents the variation with time $t$ of the kinetic energy $E_k$ of the ball? A. A straight line through the origin with positive gradient. B. A parabola opening upwards starting from the origin. C. A horizontal straight line. D. A straight line with negative gradient.

4. An electric motor has an input power of $200 \, \text{W}$ and an efficiency of $75\%$ . What is the rate at which energy is dissipated as heat and sound? A. $50 \, \text{W}$ B. $150 \, \text{W}$ C. $200 \, \text{W}$ D. $267 \, \text{W}$

5. A force $\vec{F}$ acts on an object causing a displacement $\vec{s}$ . The angle between the force and the displacement is $\theta$ . Which expression gives the work done by the force? A. $Fs \sin \theta$ B. $Fs \cos \theta$ C. $Fs \tan \theta$ D. $Fs$

Section B: Structured Calculations (Questions 6–15)

Marks vary as indicated

6. A car of mass $1200 \, \text{kg}$ accelerates from rest to a speed of $25 \, \text{m s}^{-1}$ in $10 \, \text{s}$ . (a) Calculate the gain in kinetic energy of the car. [2]

(b) Determine the average power developed by the engine during this acceleration, assuming no energy losses to resistance. [2]

7. A block of mass $4.0 \, \text{kg}$ is pulled up a smooth inclined plane by a constant force of $30 \, \text{N}$ parallel to the slope. The plane is inclined at $30^\circ$ to the horizontal. The block moves a distance of $5.0 \, \text{m}$ up the slope. (a) Calculate the work done by the pulling force. [1]

(b) Calculate the gain in gravitational potential energy of the block. [2]

8. An elevator system lifts a total mass of $800 \, \text{kg}$ through a vertical height of $15 \, \text{m}$ in $20 \, \text{s}$ . The motor driving the elevator has an input power of $8.0 \, \text{kW}$ . (a) Calculate the useful output power of the motor. [2]

(b) Calculate the efficiency of the elevator system. [2]

9. A pump raises water from a well. The water is lifted vertically through a height of $8.0 \, \text{m}$ at a rate of $0.050 \, \text{m}^3$ per minute. The density of water is $1000 \, \text{kg m}^{-3}$ . (a) Calculate the mass of water raised per second. [2]

(b) Calculate the minimum power required to raise this water. [2]

10. A toy car of mass $0.20 \, \text{kg}$ travels along a horizontal track. It enters a vertical loop of radius $0.40 \, \text{m}$ . (a) State the condition required for the car to just complete the loop at the highest point. [1]

(b) Calculate the minimum speed the car must have at the highest point of the loop to maintain contact with the track. [2]

11. A cyclist travels at a constant speed of $8.0 \, \text{m s}^{-1}$ on a level road. The total resistive force acting on the cyclist and bicycle is $45 \, \text{N}$ . (a) Explain why the cyclist must continue to pedal to maintain constant speed. [1]

(b) Calculate the power output of the cyclist. [2]

12. A spring obeys Hooke's Law. When a force of $10 \, \text{N}$ is applied, the extension is $4.0 \, \text{cm}$ . (a) Calculate the spring constant $k$ . [2]

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

13. A roller coaster car of mass $500 \, \text{kg}$ starts from rest at the top of a hill of height $30 \, \text{m}$ . It descends to the bottom of the hill. Assume friction and air resistance are negligible. (a) Calculate the speed of the car at the bottom of the hill. [3]

(b) In reality, the speed at the bottom is $22 \, \text{m s}^{-1}$ . Calculate the work done against resistive forces during the descent. [2]

14. A hydroelectric power station uses water falling from a height of $120 \, \text{m}$ to drive turbines. The flow rate of water is $200 \, \text{kg s}^{-1}$ . The efficiency of the conversion from gravitational potential energy to electrical energy is $85\%$ . (a) Calculate the gravitational potential energy lost by the water per second. [2]

(b) Calculate the electrical power output of the station. [2]

15. A box of mass $10 \, \text{kg}$ is pushed across a rough horizontal floor by a horizontal force of $50 \, \text{N}$ . The box accelerates at $2.0 \, \text{m s}^{-2}$ . (a) Calculate the net force acting on the box. [1]

(b) Calculate the frictional force acting on the box. [2]

Section C: Data Interpretation & 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 it, 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. Two cars, A and B, collide. Car A has mass $1000 \, \text{kg}$ and velocity $20 \, \text{m s}^{-1}$ . Car B has mass $1500 \, \text{kg}$ and is stationary. After the collision, they stick together. (a) Calculate the common velocity after the collision. [3]

(b) Calculate the loss in kinetic energy during the collision. [2]

18. A student investigates the power output of a small electric motor by using it to lift a mass $m$ through a height $h$ in time $t$ . (a) Write down the expression for the useful power output $P_{out}$ in terms of $m, g, h,$ and $t$ . [1]

(b) The student measures: $m = 0.50 \pm 0.01 \, \text{kg}$ $h = 1.00 \pm 0.01 \, \text{m}$ $t = 2.0 \pm 0.1 \, \text{s}$ Determine the percentage uncertainty in the calculated power output. [2]

19. A car engine produces a constant driving force. As the car speeds up, the air resistance increases. (a) Describe and explain the variation of the car's acceleration as its speed increases. [2]

(b) Explain why the power output of the engine must increase to maintain a constant acceleration as speed increases. [2]

20. A pendulum bob of mass $0.2 \, \text{kg}$ is released from a height of $0.1 \, \text{m}$ above its lowest point. (a) Calculate the maximum speed of the bob. [2]

(b) In a real experiment, the bob does not return to the original height. Explain the energy transformations that occur during one complete oscillation. [2]

Answers

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

1. B

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

2. C

Force required to lift at constant speed $F = mg = 500 \times 9.81 = 4905 \, \text{N}$ .
Power $P = Fv = 4905 \times 2.0 = 9810 \, \text{W}$ .

3. B

$v = gt$ . $E_k = \frac{1}{2}mv^2 = \frac{1}{2}m(gt)^2 = \frac{1}{2}mg^2t^2$ .
$E_k \propto t^2$ , so the graph is a parabola opening upwards.

4. A

Useful Power $P_{out} = 0.75 \times 200 = 150 \, \text{W}$ .
Dissipated Power $P_{loss} = P_{in} - P_{out} = 200 - 150 = 50 \, \text{W}$ .

5. B

Work done $W = Fs \cos \theta$ , where $\theta$ is the angle between force and displacement vectors.

6. (a) Gain in KE $= \frac{1}{2}mv^2 - 0 = \frac{1}{2}(1200)(25)^2$ [M1] $= 375,000 \, \text{J}$ or $375 \, \text{kJ}$ [A1]

(b) Average Power $P = \frac{\text{Work Done}}{\text{Time}} = \frac{\Delta KE}{t}$ [M1] $P = \frac{375,000}{10} = 37,500 \, \text{W}$ or $37.5 \, \text{kW}$ [A1]

7. (a) Work done by pulling force $W = Fs = 30 \times 5.0 = 150 \, \text{J}$ [A1]

(b) Vertical height gained $h = s \sin 30^\circ = 5.0 \times 0.5 = 2.5 \, \text{m}$ [M1] Gain in GPE $= mgh = 4.0 \times 9.81 \times 2.5 = 98.1 \, \text{J}$ [A1]

(c) By Conservation of Energy: Work Done by Force = Gain in GPE + Gain in KE [M1] $150 = 98.1 + KE_{final}$ $KE_{final} = 150 - 98.1 = 51.9 \, \text{J}$ [A1]

8. (a) Useful Output Power $P_{out} = \frac{mgh}{t}$ [M1] $P_{out} = \frac{800 \times 9.81 \times 15}{20} = 5886 \, \text{W}$ [A1] (Accept $5890 \, \text{W}$ or $5.89 \, \text{kW}$ )

(b) Efficiency $= \frac{P_{out}}{P_{in}} \times 100\%$ [M1] $P_{in} = 8.0 \, \text{kW} = 8000 \, \text{W}$ Efficiency $= \frac{5886}{8000} \times 100\% = 73.6\%$ [A1]

9. (a) Volume per second $= \frac{0.050}{60} \, \text{m}^3 \text{s}^{-1}$ [M1] Mass per second $\dot{m} = \rho \times \text{Volume rate} = 1000 \times \frac{0.050}{60} = 0.833 \, \text{kg s}^{-1}$ [A1]

(b) Minimum Power $P = \dot{m}gh$ [M1] $P = 0.833 \times 9.81 \times 8.0 = 65.4 \, \text{W}$ [A1]

10. (a) Condition: The centripetal force is provided entirely by the weight of the car (normal reaction force is zero). [B1]

(b) At highest point: $mg = \frac{mv^2}{r}$ [M1] $v^2 = gr = 9.81 \times 0.40$ $v = \sqrt{3.924} = 1.98 \, \text{m s}^{-1}$ [A1]

11. (a) To balance the resistive forces. If the cyclist stops pedalling, the net force would be resistive, causing deceleration. To maintain constant speed (zero acceleration), the driving force must equal the resistive force. [B1]

(b) Driving Force $F = \text{Resistive Force} = 45 \, \text{N}$ [M1] Power $P = Fv = 45 \times 8.0 = 360 \, \text{W}$ [A1]

12. (a) Hooke's Law: $F = kx$ [M1] $k = \frac{F}{x} = \frac{10}{0.04} = 250 \, \text{N m}^{-1}$ [A1]

(b) Elastic Potential Energy $E_p = \frac{1}{2}Fx$ or $\frac{1}{2}kx^2$ [M1] $E_p = \frac{1}{2} \times 10 \times 0.04 = 0.20 \, \text{J}$ [A1]

13. (a) Conservation of Energy: Loss in GPE = Gain in KE [M1] $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ [M1] $v = \sqrt{2 \times 9.81 \times 30} = \sqrt{588.6} = 24.3 \, \text{m s}^{-1}$ [A1]

(b) Actual KE $= \frac{1}{2}(500)(22)^2 = 121,000 \, \text{J}$ [M1] Initial GPE $= 500 \times 9.81 \times 30 = 147,150 \, \text{J}$ Work done against resistance $= \text{Initial GPE} - \text{Final KE}$ $W = 147,150 - 121,000 = 26,150 \, \text{J}$ [A1] (Accept $26 \, \text{kJ}$ )

14. (a) GPE lost per second $= \dot{m}gh$ [M1] $P_{input} = 200 \times 9.81 \times 120 = 235,440 \, \text{W}$ [A1] (Accept $235 \, \text{kW}$ )

(b) Electrical Power Output $= \text{Efficiency} \times P_{input}$ [M1] $P_{out} = 0.85 \times 235,440 = 200,124 \, \text{W}$ [A1] (Accept $200 \, \text{kW}$ )

15. (a) Net Force $F_{net} = ma = 10 \times 2.0 = 20 \, \text{N}$ [A1]

(b) $F_{net} = F_{push} - F_{friction}$ [M1] $20 = 50 - F_{friction}$ $F_{friction} = 30 \, \text{N}$ [A1]

16. (a) The work done in stretching the rubber band (or energy stored). [B1]

(b) The area represents the energy dissipated as heat (internal energy) during the loading-unloading cycle. This is due to hysteresis. [B1] for "energy dissipated/lost as heat", [B1] for "hysteresis" or explanation of internal friction.

17. (a) Conservation of Momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ [M1] $(1000)(20) + 0 = (1000 + 1500)v$ $20,000 = 2500v$ $v = 8.0 \, \text{m s}^{-1}$ [A1]

(b) Initial KE $= \frac{1}{2}(1000)(20)^2 = 200,000 \, \text{J}$ [M1] Final KE $= \frac{1}{2}(2500)(8.0)^2 = 80,000 \, \text{J}$ Loss in KE $= 200,000 - 80,000 = 120,000 \, \text{J}$ [A1]

18. (a) $P_{out} = \frac{mgh}{t}$ [B1]

(b) Percentage uncertainty in $P = \%u_m + \%u_h + \%u_t$ [M1] $\%u_m = \frac{0.01}{0.50} \times 100 = 2\%$ $\%u_h = \frac{0.01}{1.00} \times 100 = 1\%$ $\%u_t = \frac{0.1}{2.0} \times 100 = 5\%$ Total $\%u = 2 + 1 + 5 = 8\%$ [A1]

19. (a) Acceleration decreases. [B1] As speed increases, air resistance increases, so the resultant force ( $F_{drive} - F_{resistance}$ ) decreases. Since $a = F_{net}/m$ , acceleration decreases. [B1]

(b) Power $P = Fv$ . [B1] To maintain constant acceleration, the net force must be constant. Since resistive force increases with speed, the driving force $F_{drive}$ must increase. Since both $F_{drive}$ and $v$ increase, the power output $P$ must increase. [B1]

20. (a) Conservation of Energy: $mgh = \frac{1}{2}mv^2$ [M1] $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.1} = \sqrt{1.962} = 1.40 \, \text{m s}^{-1}$ [A1]

(b) GPE converts to KE as it falls. [B1] At the bottom, KE is maximum. As it rises, KE converts back to GPE. However, some energy is dissipated as heat/sound due to air resistance and friction at the pivot, so it does not reach the original height. [B1]