JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Work, Power & Energy

A particle of mass 500 gm is moving in a straight line with velocity

v = b x^{5/2}

. The work done by the net force during its displacement from

x = 0

x = 4

m is: (Take

b = 0.25 m^{- 3/2} s^{- 1}

(A)2 J
(B)4 J
(C)8 J
(D)16 J✓ Correct

Answer: (D)

Solution

By the work-energy theorem, work done equals the change in kinetic energy:

W = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})

. At

x = 0

v_{i} = 0

. At

x = 4

v_{f} = 0.25 \times (4)^{5/2} = 0.25 \times 32 = 8

m/s. Thus

W = \frac{1}{2} \times 0.5 \times (8^{2} - 0) = 0.25 \times 64 = 16

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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