A body of mass 0.5 kg travels on straight line path with velocity $\mathrm{v}=(3 \mathrm{x}^{2}+4) \mathrm{m} / \mathrm{s}$. The net workdone by the force during its displacement from $\mathrm{x}=0$ to $\mathrm{x}=2 \mathrm{~m}$ is :

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Work, Power & Energy

A body of mass 0.5 kg travels on straight line path with velocity

v = (3 x^{2} + 4) m / s

. The net workdone by the force during its displacement from

x = 0

x = 2 m

is :

(A)64 J
(B)60 J✓ Correct
(C)120 J
(D)128 J

Answer: (B)

Solution

Using the work-energy theorem, the net work done equals the change in kinetic energy.

At

x = 0

v_{i} = 3 (0)^{2} + 4 = 4

m/s

At

x = 2

v_{f} = 3 (2)^{2} + 4 = 16

m/s

Work done:

W = Δ K E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (0.5) (1 6^{2} - 4^{2})

= \frac{1}{4} (256 - 16) = \frac{240}{4} = 60 J

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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