A Level H1 Physics Energy Power Quiz

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]