Potential Energy — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Potential Energy MCQs with step-by-step solutions (8 questions). Part of Work, Power and Energy. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Potential Energy · easy · theory
Potential energy can be defined only for:
A. Contact forces only
B. Constant forces only
C. Conservative forces ✓ Correct
D. All forces including friction
Solution: U is defined through ΔU = −W_cons, which requires path-independent (conservative) forces. Its zero level is an arbitrary choice.
Q2 — Potential Energy · medium · theory
A spring is compressed by x, and separately stretched by the same x. The elastic potential energies stored are:
A. Greater for the compression
B. Equal (½kx² in both cases) ✓ Correct
C. Greater for the stretch
D. Zero for compression
Solution: U = ½kx² depends on x², so equal deformation stores equal energy either way.
Q3 — Potential Energy · easy · numerical
A 2 kg block is raised through a height of 15 m. The gain in its potential energy is:
A. 300 J ✓ Correct
B. 30 J
C. 600 J
D. 150 J
Solution: ΔU = mgh = 2×10×15 = 300 J.
Q4 — Potential Energy · medium · numerical
A spring of k = 100 N/m is stretched by 20 cm. The potential energy stored is:
A. 0.2 J
B. 2 J ✓ Correct
C. 4 J
D. 20 J
Solution: U = ½kx² = ½×100×(0.2)² = 50×0.04 = 2 J. Trap: convert cm to m before squaring.
Q5 — Potential Energy · hard · numerical
A uniform rod of mass 2 kg and length 1 m lies flat on the ground. The work needed to stand it vertically on one end is:
A. 5 J
B. 10 J ✓ Correct
C. 20 J
D. 40 J
Solution: Only the centre of mass matters: it rises by L/2 = 0.5 m. W = mg(L/2) = 2×10×0.5 = 10 J.
Q6 — Potential Energy · medium · numerical
A spring stores 4 J of energy when stretched by x. Keeping the same spring, the energy stored at stretch 2x is:
A. 8 J
B. 16 J ✓ Correct
C. 12 J
D. 4 J
Solution: U ∝ x² ⇒ ×4 at double stretch: 16 J.
Q7 — Potential Energy · hard · numerical
A spring stores 25 J at stretch x. The ADDITIONAL work required to stretch it from x to 2x is:
A. 50 J
B. 75 J ✓ Correct
C. 100 J
D. 25 J
Solution: U(2x) = 4×25 = 100 J. Additional work = 100 − 25 = 75 J. Trap: not 25 J — the force grows with stretch.
Q8 — Potential Energy · medium · numerical
The potential energy of a particle is U(x) = 20 + (x − 2)² (J). Its minimum potential energy, and the position where the force vanishes, are:
A. 20 J at x = 0
B. 24 J at x = 0
C. 20 J at x = 2 m ✓ Correct
D. Zero at x = 2 m
Solution: (x−2)² ≥ 0, so U_min = 20 J at x = 2 m; F = −dU/dx = −2(x−2) = 0 there (stable equilibrium). Trap: U_min need not be zero.