Work, Energy and Power

8 NEET-UG practice questions on Work, Energy and Power, part of Physics. The first 8 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   A force F = (3x² + 2) N, with x in metres, acts along the x-axis on a 1.5 kg block resting at x = 0 on a smooth surface. What is the speed of the block at x = 2 m?

   1. A4 m/s
   2. B16 m/s
   3. C6.1 m/s
   4. D4.6 m/s

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   Answer: A · 4 m/s

   Work = integral of F dx from 0 to 2 = [x³ + 2x] = 8 + 4 = 12 J. By the work-energy theorem, (1/2)(1.5)v² = 12, so v² = 16 and v = 4 m/s.

   • B: This gives v², forgetting the square root.
   • C: This uses the final force of 14 N over the whole 2 m.
   • D: This takes the mean of the force at the two ends, which fails for a non-linear force.

   Remember this

   For a variable force, work is the integral of F dx, then use the work-energy theorem.

2. Question 2 · Hard

   A 1 kg particle moves along the x-axis under a conservative force with potential energy U(x) = (2x² − 8x + 10) J, x in metres. Its total mechanical energy is 10 J. What are its maximum speed and the position where this occurs?

   1. A2√2 m/s at x = 2 m
   2. B√20 m/s at x = 2 m
   3. C4 m/s at x = 4 m
   4. D4 m/s at x = 2 m

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   Answer: D · 4 m/s at x = 2 m

   U is least where dU/dx = 4x − 8 = 0, so at x = 2 m, where U = 2 J. Then KE = 10 − 2 = 8 J, so (1/2)(1)v² = 8 and v = 4 m/s. Check: this is SHM with k = 4 N/m, ω = 2 rad/s and turning points x = 0 and 4 m (A = 2 m), so v = Aω = 4 m/s.

   • A: This forgets the factor 1/2 in kinetic energy.
   • B: This takes the minimum of U to be zero rather than 2 J.
   • C: x = 4 m is a turning point, where the speed is zero.

   Remember this

   Speed is greatest where potential energy is least; turning points lie where U equals the total energy.

3. Question 3 · Easy

   In 60 s, a pump lifts 300 kg of water through a height of 20 m. What is the minimum power it needs? Take g = 10 m/s².

   1. A50 W
   2. B1000 W
   3. C60000 W
   4. D100 W

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   Answer: B · 1000 W

   Work against gravity = mgh = 300 × 10 × 20 = 60000 J. Power = 60000/60 = 1000 W.

   • A: This leaves out the height in the work.
   • C: This is the work in joules, not divided by the time.
   • D: This leaves out g in the work.

   Remember this

   Power is the rate of doing work, so always divide the work by the time taken.

4. Question 4 · Medium

   A 0.5 kg stone on a 1 m string is whirled in a vertical circle, with just enough speed at the top to keep the string taut. Taking g = 10 m/s², what is the tension in the string at the lowest point?

   1. A30 N
   2. B15 N
   3. C25 N
   4. D5 N

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   Answer: A · 30 N

   At the top, tension is zero and v² = gL = 10. Energy conservation gives v² at the bottom = 10 + 4gL = 50. At the bottom, T = mg + mv²/L = 5 + 25 = 30 N, which is 6mg.

   • B: This uses v² = 2gL at the bottom instead of 5gL.
   • C: This leaves out the weight and keeps only mv²/L.
   • D: This is the weight alone.

   Remember this

   For the minimum case, the tension at the lowest point is 6mg and the speed there is √(5gL).

5. Question 5 · Medium

   Given below are two statements: Statement I: In a perfectly inelastic collision of an isolated system, the total linear momentum is conserved but some kinetic energy is lost. Statement II: In an elastic collision, the total kinetic energy after the collision equals that before it, although kinetic energy is not conserved during the contact. Choose the correct answer.

   1. ABoth Statement I and Statement II are correct
   2. BStatement I is incorrect but Statement II is correct
   3. CBoth Statement I and Statement II are incorrect
   4. DStatement I is correct but Statement II is incorrect

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   Answer: A · Both Statement I and Statement II are correct

   Linear momentum is conserved in every collision of an isolated system, while a perfectly inelastic collision loses the maximum kinetic energy consistent with this. In an elastic collision, some kinetic energy turns into deformation energy during contact and returns afterwards. So both statements are correct.

   • B: Statement I is correct: momentum is conserved but kinetic energy is not.
   • C: Both statements match the definitions of the two collision types.
   • D: Statement II is correct: kinetic energy is stored briefly during contact.

   Remember this

   Elastic means initial and final kinetic energies are equal, not that kinetic energy is constant at every instant.

6. Question 6 · Medium

   Given below are two statements: Statement I: On the potential energy curve shown, point R is a position of stable equilibrium. Statement II: A particle moving under this potential energy alone, released from rest at P, reaches S with non-zero kinetic energy. Choose the correct answer.

   

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   1. AStatement I is incorrect but Statement II is correct
   2. BBoth Statement I and Statement II are correct
   3. CStatement I is correct but Statement II is incorrect
   4. DBoth Statement I and Statement II are incorrect

   Answer and explanation in LearnFree account

7. Question 7 · Hard

   A small 1 kg block is released from rest at the top of the frictionless track, which ends in a vertical circular loop. With g = 10 m/s², what normal reaction does the track exert on the block at the top T of the loop?

   

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   1. A10 N
   2. B20 N
   3. C0 N
   4. D30 N

   Answer and explanation in LearnFree account

8. Question 8 · Easy

   Force along the direction of motion varies with position x as plotted. Calculate the work done by this force from x = 0 to x = 8 m.

   

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   1. A20 J
   2. B60 J
   3. C40 J
   4. D80 J

   Answer and explanation in LearnFree account