Work and Energy

Practise Work and Energy for FBISE Grade 9–10 Physics (Grade 9–10): 8 multiple-choice questions with instant checking and worked explanations. Free, no sign-up.

Tap an answer to check it · 8 questions

1. Question 1

   What is the SI unit of work?
   • AWatt
   • BPascal
   • CJoule
   • DNewton
   Show answer and explanation

   Answer: C) Joule

   The joule (J) is the SI unit of work; one joule equals one newton‑metre. The watt measures power (energy per unit time) and is therefore not the unit of work.

2. Question 2

   Which physical quantity is defined as the rate of doing work?
   • AMomentum
   • BPower
   • CEnergy
   • DTorque
   Show answer and explanation

   Answer: B) Power

   Power is defined as the rate at which work is done, i.e., work per unit time. Energy is the amount of work done, not its rate. Momentum measures motion and is unrelated to work or its rate. Torque is a rotational force, not the time derivative of work.

3. Question 3

   In which of the following situations does the specified force do negative work on the moving object?
   • AGravity acting on a freely falling stone
   • BTension in a cable lifting an elevator upward
   • CFriction acting on a skidding bicycle tyre
   • DNormal force supporting a box sliding on a floor
   Show answer and explanation

   Answer: C) Friction acting on a skidding bicycle tyre

   Work is positive when the force component along the displacement points in the same direction as the motion, negative when opposite, and zero when perpendicular. Friction on a skidding tyre opposes the tyre’s motion, so it does negative work. Tension on an elevator and gravity on a falling stone act in the same direction as the displacement, giving positive work. The normal force is perpendicular to the sliding motion, so its work is zero.

4. Question 4

   Two identical trolleys, X and Y, move on a frictionless surface such that trolley X has twice the momentum of trolley Y. How does the kinetic energy of trolley X compare to that of trolley Y?
   • AIt is eight times as great
   • BIt is four times as great
   • CIt is equal
   • DIt is twice as great
   Show answer and explanation

   Answer: B) It is four times as great

   Because the two trolleys have the same mass, the ratio of their momenta equals the ratio of their speeds. If trolley X has twice the momentum of trolley Y, its speed is twice as large. Kinetic energy depends on the square of speed, so $K_X=\frac{1}{2}m(2v{)}^2=4\frac{1}{2}m{v}^2=4K_Y$. The other options stem from treating kinetic energy as linear with momentum or ignoring the speed dependence.

5. Question 5

   An electric crane raises a $400\mathrm{kg}$ load vertically at a constant speed of $1.5\mathrm{m}{\mathrm{s}}^{-1}$ while drawing $8.0\mathrm{kW}$ of electrical power. Taking $g=10\mathrm{m}{\mathrm{s}}^{-2}$, what is the efficiency of the crane?
   • A$80\%$
   • B$50\%$
   • C$60\%$
   • D$75\%$
   Show answer and explanation

   Answer: D) $75\%$

   The crane must supply enough mechanical power to lift the load at constant speed, which is $P=mgv$. Using $m=400\mathrm{kg}$, $g=10\mathrm{m}{\mathrm{s}}^{-2}$, and $v=1.5\mathrm{m}{\mathrm{s}}^{-1}$ gives $6\mathrm{kW}$. Comparing this output power with the electrical input of $8\mathrm{kW}$ yields an efficiency of $6/8=75\%$. The other options arise from mixing up work and power or mis‑interpreting the conversion of gravitational energy.

6. Question 6

   A $20\mathrm{g}$ bullet travelling horizontally at $300\mathrm{m}{\mathrm{s}}^{-1}$ strikes a wooden block and penetrates $15\mathrm{cm}$ before coming to rest. What average retarding force is exerted by the wood on the bullet?
   • A$6.0\mathrm{kN}$
   • B$1.8\mathrm{kN}$
   • C$12.0\mathrm{kN}$
   • D$600\mathrm{N}$
   Show answer and explanation

   Answer: A) $6.0\mathrm{kN}$

   According to the work-energy theorem, the work done by the retarding force equals the initial kinetic energy of the bullet. First convert mass to kilograms ($20\mathrm{g}=0.020\mathrm{kg}$) and distance to metres ($15\mathrm{cm}=0.15\mathrm{m}$). The kinetic energy lost by the bullet is $E_k=\frac{1}{2}m{v}^2=\frac{1}{2}(0.020\mathrm{kg})(300\mathrm{m}{\mathrm{s}}^{-1}{)}^2=900\mathrm{J}$. Equating this to work done $W=F\times d$ gives $900\mathrm{J}=F\times 0.15\mathrm{m}$, which yields $F=6000\mathrm{N}=6.0\mathrm{kN}$.

7. Question 7

   A $1000\mathrm{kg}$ car accelerates uniformly from rest to a speed of $20\mathrm{m}{\mathrm{s}}^{-1}$ in $10\mathrm{s}$ along a straight horizontal road. Throughout the motion, a constant resistive force of $500\mathrm{N}$ opposes the car. What average power is delivered by the engine during this period?
   • A$20\mathrm{kW}$
   • B$15\mathrm{kW}$
   • C$30\mathrm{kW}$
   • D$25\mathrm{kW}$
   Show answer and explanation

   Answer: D) $25\mathrm{kW}$

   The engine must supply work both to increase the car's kinetic energy and to overcome resistance. The distance travelled during uniform acceleration from rest is $d=\frac{1}{2}(v_i+v_f)t=\frac{1}{2}(0+20\mathrm{m}{\mathrm{s}}^{-1})(10\mathrm{s})=100\mathrm{m}$. The gain in kinetic energy is $\Delta E_k=\frac{1}{2}m{v}_f^2=\frac{1}{2}(1000\mathrm{kg})(20\mathrm{m}{\mathrm{s}}^{-1}{)}^2=200,000\mathrm{J}$. The work done against the resistive force is $W_r=F_{r}d=500\mathrm{N}\times 100\mathrm{m}=50,000\mathrm{J}$. Thus, total engine work is $250,000\mathrm{J}$. Dividing total work by $10\mathrm{s}$ gives an average power of $25,000\mathrm{W}=25\mathrm{kW}$.

8. Question 8

   A $4\mathrm{kg}$ block starts from rest and slides $10\mathrm{m}$ down a rough incline of $30^{\circ }$ against a constant frictional force of $4\mathrm{N}$. At the bottom, it moves along a frictionless horizontal surface and compresses a spring with a force constant of $80\mathrm{N}{\mathrm{m}}^{-1}$. Taking $g=10\mathrm{m}{\mathrm{s}}^{-2}$, what is the maximum compression of the spring?
   • A$2.2\mathrm{m}$
   • B$2.0\mathrm{m}$
   • C$1.6\mathrm{m}$
   • D$4.0\mathrm{m}$
   Show answer and explanation

   Answer: B) $2.0\mathrm{m}$

   By conservation of energy, the mechanical energy stored in the compressed spring equals the initial gravitational potential energy minus the work done against friction during the slide down the incline. The vertical drop is $h=(10\mathrm{m})\sin 30^{\circ }=5\mathrm{m}$, giving an initial potential energy of $mgh=4\mathrm{kg}\times 10\mathrm{m}{\mathrm{s}}^{-2}\times 5\mathrm{m}=200\mathrm{J}$. The work done against friction is $W_f=F_f\times d=4\mathrm{N}\times 10\mathrm{m}=40\mathrm{J}$. Thus, the energy available to compress the spring is $200\mathrm{J}-40\mathrm{J}=160\mathrm{J}$. Setting this equal to elastic potential energy $\frac{1}{2}k{x}^2=160\mathrm{J}$ with $k=80\mathrm{N}{\mathrm{m}}^{-1}$ yields $40x^2=160$, so $x=2.0\mathrm{m}$.

More FBISE Grade 9–10 Physics topics

• Dynamics (Newton's Laws) (8)
• Gravitation (8)
• Kinematics (5)
• Simple Machines (8)
• Physical Quantities & Measurement (6)
• Properties of Matter (8)