A Level H2 Physics Energy Power Quiz

A-Level Physics H2 Quiz - Energy Power (Answer Key)

1. A
[1]

2. B
$P = Fv = 800 \times 25 = 20,000 \text{ W} = 20 \text{ kW}$.
[1]

3. Energy cannot be created or destroyed; it can only be transformed from one form to another or transferred from one body to another. The total energy of an isolated system remains constant.
[2] (1 mark for "cannot be created/destroyed", 1 mark for "transformed/transferred" or "total constant")

4. Gravitational potential energy is converted into kinetic energy and internal energy (heat) due to work done against air resistance. As terminal velocity is reached, the rate of loss of GPE equals the rate of work done against air resistance, so KE remains constant.
[2] (1 mark for GPE to KE + Internal/Heat, 1 mark for mention of terminal velocity condition or constant KE)

5. Power is the rate of doing work or the rate of energy transfer.
[1]

6.
(a) $v = u + at = 0 + (0.50)(4.0) = 2.0 \text{ m s}^{-1}$.
[1]

(b) Height $h = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ m}$.
$\Delta E_p = mgh = 500 \times 9.81 \times 4.0 = 19,620 \text{ J} \approx 19.6 \text{ kJ}$.
[2] (1 mark for height, 1 mark for correct energy)

(c) $T - mg = ma \Rightarrow T = m(g+a) = 500(9.81 + 0.50) = 500(10.31) = 5155 \text{ N} \approx 5.2 \text{ kN}$.
[2] (1 mark for equation, 1 mark for answer)

(d) Work Done by Tension $W = T \times h = 5155 \times 4.0 = 20,620 \text{ J}$.
Average Power $P = \frac{W}{t} = \frac{20,620}{4.0} = 5155 \text{ W} \approx 5.2 \text{ kW}$.
Alternative: Average velocity $\bar{v} = \frac{0+2}{2} = 1.0 \text{ m s}^{-1}$. $P = T \bar{v} = 5155 \times 1.0 = 5.2 \text{ kW}$.
[3] (1 mark for Work/Energy or Force x Avg Vel, 1 mark for substitution, 1 mark for answer)

7.
(a) $P_{out} = \text{Efficiency} \times P_{in} = 0.60 \times 2500 = 1500 \text{ W}$.
[1]

(b) Useful Power $P = \frac{mgh}{t}$.
$1500 = \frac{m \times 9.81 \times 12}{60}$ (time = 60s for "per minute").
$m = \frac{1500 \times 60}{9.81 \times 12} = \frac{90000}{117.72} \approx 764.5 \text{ kg}$.
Answer: $760 \text{ kg}$ (2 s.f.).
[3] (1 mark for formula rearrangement, 1 mark for substitution, 1 mark for answer)

8.
(a) Vertical height $h = 5.0 \sin(30^\circ) = 2.5 \text{ m}$.
$\Delta E_p = mgh = 2.0 \times 9.81 \times 2.5 = 49.05 \text{ J} \approx 49 \text{ J}$.
[2] (1 mark for height, 1 mark for energy)

(b) $\Delta E_k = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$.
[1]

(c) Work done against friction $W_f = \Delta E_p - \Delta E_k = 49.05 - 16 = 33.05 \text{ J}$.
$W_f = F_f \times d \Rightarrow 33.05 = F_f \times 5.0$.
$F_f = \frac{33.05}{5.0} = 6.61 \text{ N} \approx 6.6 \text{ N}$.
[3] (1 mark for energy difference, 1 mark for work-force relation, 1 mark for answer)

9.
(a) At constant velocity, Driving Force = Resistive Force.
$3000 = k(20)^2 \Rightarrow 3000 = 400k \Rightarrow k = \frac{3000}{400} = 7.5 \text{ N s}^2 \text{ m}^{-2}$ (or $\text{kg m}^{-1}$).
[2] (1 mark for equilibrium condition, 1 mark for k)

(b) At $v = 10 \text{ m s}^{-1}$, $F_{res} = 7.5(10)^2 = 750 \text{ N}$.
Net Force $F_{net} = F_{drive} - F_{res} = 3000 - 750 = 2250 \text{ N}$.
$a = \frac{F_{net}}{m} = \frac{2250}{1500} = 1.5 \text{ m s}^{-2}$.
[3] (1 mark for new resistive force, 1 mark for net force, 1 mark for acceleration)

10.
(a) Mass per second $\dot{m} = \text{Density} \times \text{Flow Rate} = 1000 \times 200 = 200,000 \text{ kg s}^{-1}$.
[1]

(b) Input Power (Gravitational) $P_{in} = \dot{m}gh = 200,000 \times 9.81 \times 150 = 294,300,000 \text{ W}$.
Output Power $P_{out} = 0.85 \times P_{in} = 0.85 \times 294,300,000 = 250,155,000 \text{ W}$.
Answer: $250 \text{ MW}$ (or $2.5 \times 10^8 \text{ W}$).
[3] (1 mark for input power calc, 1 mark for efficiency application, 1 mark for final answer)

11.
(a) Component of weight $= mg \sin(\theta) = 80 \times 9.81 \times \sin(5.0^\circ)$.
$\sin(5.0^\circ) \approx 0.08716$.
$W_{parallel} = 80 \times 9.81 \times 0.08716 \approx 68.4 \text{ N}$.
[2] (1 mark for formula, 1 mark for answer)

(b) Total force required $F = W_{parallel} + F_{resistive} = 68.4 + 40 = 108.4 \text{ N}$.
Power $P = Fv = 108.4 \times 8.0 = 867.2 \text{ W} \approx 870 \text{ W}$.
[3] (1 mark for total force, 1 mark for P=Fv, 1 mark for answer)

12.
(a) $\log P = 2 \log I + 0.60 \Rightarrow P = 10^{0.60} I^2$.
Relationship: $P \propto I^2$.
[2] (1 mark for identifying power law, 1 mark for stating proportionality)

(b) Comparing to $P = I^2 R$, the constant $k = 10^{0.60} = R$.
$R = 10^{0.60} \approx 3.98 \Omega \approx 4.0 \Omega$.
[2] (1 mark for identifying intercept as log R, 1 mark for calculation)

13.
Thrust force $F$ is constant. Work done by thrust over distance $d$ is $W = Fd$.
As mass $m$ decreases, for the same force $F$, acceleration $a = F/m$ increases.
From energy perspective: The rate of change of kinetic energy ($P = Fv$) leads to a faster increase in velocity $v$ as mass drops, because $KE = \frac{1}{2}mv^2$. With lower $m$, a smaller increase in energy yields a larger increase in $v$, implying higher acceleration.
Acceptable Answer: $F=ma$. Since $F$ is constant and $m$ decreases, $a$ must increase. The work done by the engine goes into increasing KE. As mass drops, the same work results in a larger change in velocity.
[3] (1 mark for F=ma or Work-Energy link, 1 mark for mass decrease effect, 1 mark for conclusion on acceleration)

14.
(a) Conservation of Energy for bob: $mgh = \frac{1}{2}mv^2$.
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.20} = \sqrt{3.924} \approx 1.98 \text{ m s}^{-1}$.
[2] (1 mark for formula, 1 mark for answer)

(b) Conservation of Momentum for collision: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v_{final}$.
$(0.50)(1.98) + 0 = (0.50 + 1.50)v_{final}$.
$0.99 = 2.0 v_{final} \Rightarrow v_{final} = 0.495 \text{ m s}^{-1}$.
$KE_{final} = \frac{1}{2}(m_1+m_2)v_{final}^2 = \frac{1}{2}(2.0)(0.495)^2 \approx 0.245 \text{ J}$.
[3] (1 mark for momentum conservation to find v, 1 mark for v value, 1 mark for final KE)

15.
(a) Total Solar Power Incident $= 800 \times 20 = 16,000 \text{ W}$.
Electrical Power $= 0.15 \times 16,000 = 2,400 \text{ W} = 2.4 \text{ kW}$.
[2] (1 mark for incident power, 1 mark for efficiency calc)

(b) Daily Energy Demand $= 10 \text{ kWh} = 10 \times 1000 \text{ Wh} = 10,000 \text{ Wh}$.
Power generated $= 2.4 \text{ kW}$.
Time $t = \frac{\text{Energy}}{\text{Power}} = \frac{10 \text{ kWh}}{2.4 \text{ kW}} \approx 4.17 \text{ hours}$.
Answer: $4.2 \text{ hours}$.
[3] (1 mark for unit consistency/conversion, 1 mark for formula, 1 mark for answer)

16.
(a) Normal reaction $N = mg = 10 \times 9.81 = 98.1 \text{ N}$.
Frictional force $F_f = \mu N = 0.30 \times 98.1 = 29.43 \text{ N} \approx 29 \text{ N}$.
[2] (1 mark for normal force, 1 mark for friction calc)

(b) Work done by student $W = F d = 50 \times 5.0 = 250 \text{ J}$.
[1]

(c) Net force $F_{net} = F_{applied} - F_f = 50 - 29.43 = 20.57 \text{ N}$.
Net Work $W_{net} = F_{net} d = 20.57 \times 5.0 = 102.85 \text{ J} \approx 100 \text{ J}$ (2 s.f.).
Alternative: $W_{net} = W_{applied} - W_{friction} = 250 - (29.43 \times 5) = 250 - 147.15 = 102.85 \text{ J}$.
[2] (1 mark for net force or work diff, 1 mark for answer)

17.
(a) Total mass $M = 800 + 200 = 1000 \text{ kg}$.
Weight $W = Mg = 1000 \times 9.81 = 9810 \text{ N}$.
Since speed is constant, Tension $T = W + F_{friction} = 9810 + 1500 = 11,310 \text{ N} \approx 11.3 \text{ kN}$.
[2] (1 mark for weight + friction logic, 1 mark for answer)

(b) Power $P = T v = 11,310 \times 3.0 = 33,930 \text{ W} \approx 34 \text{ kW}$.
[2] (1 mark for formula, 1 mark for answer)

18.
(a) $E_p = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.15)^2 = 100 \times 0.0225 = 2.25 \text{ J}$.
[2] (1 mark for formula, 1 mark for answer)

(b) $E_k = E_p \Rightarrow \frac{1}{2}mv^2 = 2.25$.
$v^2 = \frac{2 \times 2.25}{0.05} = \frac{4.5}{0.05} = 90$.
$v = \sqrt{90} \approx 9.49 \text{ m s}^{-1} \approx 9.5 \text{ m s}^{-1}$.
[2] (1 mark for equating energies, 1 mark for answer)

19.
(a) Work done $W = mgh = 500 \times 9.81 \times 10 = 49,050 \text{ J}$.
Power $P = \frac{W}{t} = \frac{49,050}{20} = 2,452.5 \text{ W} \approx 2.45 \text{ kW}$.
[2] (1 mark for work/energy, 1 mark for power)

(b) Efficiency $= \frac{P_{useful}}{P_{input}} \times 100\% = \frac{2452.5}{5000} \times 100\% = 49.05\% \approx 49\%$.
[2] (1 mark for ratio, 1 mark for answer)

20.
(a) $P = Fv \Rightarrow F = \frac{P}{v} = \frac{100,000}{20} = 5,000 \text{ N}$.
$F = ma \Rightarrow a = \frac{F}{m} = \frac{5,000}{1,200} = 4.166... \text{ m s}^{-2} \approx 4.2 \text{ m s}^{-2}$.
[2] (1 mark for force from power, 1 mark for acceleration)

(b) Resistive forces (air resistance, friction) act against the motion, reducing the net force available for acceleration.
[1] (1 mark for mention of resistive forces)