Class 10 Work, Energy and Power Mock Test (ICSE)

Topics in Work, Energy and Power

• Work done by a force
• Power and its units
• Potential and kinetic energy
• Work-energy theorem
• Conservation of mechanical energy
• Energy conversions

Work, Energy and Power practice questions with answers

1. Q1Multiple choice·1 mark
   A constant force of $10\text{ N}$ acts on a body and displaces it by $5\text{ m}$ in a direction making an angle of $60^\circ$ with the line of action of the force. The work done by the force is:

   1. (a)$0\text{ J}$
   2. (b)$50\text{ J}$
   3. (c)$25\text{ J}$
   4. (d)$25\sqrt{3}\text{ J}$
   Show answer and step marks

   1. Correct option: (c) $25\text{ J}$+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   Which of the following units is equivalent to a kilowatt-hour (kWh)?

   1. (a)$3.6 \times 10^3\text{ J}$
   2. (b)$3.6 \times 10^4\text{ J}$
   3. (c)$3.6 \times 10^6\text{ J}$
   4. (d)$3.6 \times 10^5\text{ J}$
   Show answer and step marks

   1. Correct option: (c) $3.6 \times 10^6\text{ J}$+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   According to the work-energy theorem, the work done by the net force on a moving body is equal to change in its:

   1. (a)potential energy
   2. (b)momentum
   3. (c)kinetic energy
   4. (d)total mechanical energy
   Show answer and step marks

   1. Correct option: (c) kinetic energy+1

   Answer: (c)
4. Q4Multiple choice·1 mark
   If the linear momentum of a body is increased by $100\%$, its kinetic energy increases by:

   1. (a)$200\%$
   2. (b)$100\%$
   3. (c)$300\%$
   4. (d)$400\%$
   Show answer and step marks

   1. Correct option: (c) $300\%$+1

   Answer: (c)
5. Q5Long answer·4 marks
   Water falls from a height of $60\text{ m}$ at the rate of $500\text{ kg s}^{-1}$ to drive a turbine. (a) Calculate the total potential energy of water available per second. (b) If $40\%$ of this available energy is converted into electrical energy, find the electrical power generated. (Take $g = 10\text{ m s}^{-2}$)
   Show answer and step marks

   1. State the formula for potential energy available per second: $\text{Power} = \frac{mgh}{t} = \left(\frac{m}{t}\right)gh$+1
   2. Substitute values and calculate: $500\text{ kg s}^{-1} \times 10\text{ m s}^{-2} \times 60\text{ m} = 300,000\text{ W}$ (or $300\text{ kW}$)+1
   3. State the relation for electrical power generated: $P_{\text{elec}} = \frac{40}{100} \times \text{Total Power}$+1
   4. Calculate final electrical power: $0.40 \times 300\text{ kW} = 120\text{ kW}$ (or $120,000\text{ W}$)+1

   Answer: (a) $300000\text{ J}$ per second ($300\text{ kW}$) (b) $120\text{ kW}$ (or $120000\text{ W}$)
6. Q6Long answer·4 marks
   (a) State the energy conversion taking place in the following devices during their working: (i) An electric motor (ii) A steam engine [2] (b) A body of mass $500\text{ g}$ is thrown vertically upwards with an initial velocity of $20\text{ m s}^{-1}$. Assuming complete conversion of kinetic energy into potential energy at the highest point, calculate the maximum potential energy possessed by the body at its highest point. (Take $g = 10\text{ m s}^{-2}$) [2]
   Show answer and step marks

   1. (a) (i) Electrical energy to mechanical energy+1
   2. (a) (ii) Heat energy (thermal energy, obtained from the chemical energy of the fuel) to mechanical energy+1
   3. (b) Initial kinetic energy with mass in kg: $\text{KE} = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.5\text{ kg} \times (20\text{ m s}^{-1})^2 = 100\text{ J}$+1
   4. (b) At the highest point all the kinetic energy has become potential energy, so maximum $\text{PE} = 100\text{ J}$+1

   Answer: (a) (i) Electrical to mechanical, (ii) Heat to mechanical; (b) $100\text{ J}$

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