Work Energy Power

JEE Main level, single correct, medium

42 of 183

Question 42 of 183

A body of mass 2 kg is moving along xx-direction such that its displacement as function of time is given by x(t)=αt2+βt+γmx(t)=\alpha t^{2}+\beta t+\gamma \mathrm{m}, where α=1 m/s2,β=1 m/s\alpha=1 \mathrm{~m} / \mathrm{s}^{2}, \beta=1 \mathrm{~m} / \mathrm{s} and γ=1 m\gamma=1 \mathrm{~m}. The work done on the body during the time interval t=2 st=2 \mathrm{~s} to t=3 st=3 \mathrm{~s}, is ____\_\_\_\_ J.

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Answer: (D) 24

Given x(t)=t2+t+1x(t) = t^2 + t + 1, with m=2m = 2 kg, find work from t=2t = 2 s to t=3t = 3 s.

Velocity: v(t)=dxdt=2t+1v(t) = \frac{dx}{dt} = 2t + 1
At t=2t = 2 s: vi=2(2)+1=5v_i = 2(2) + 1 = 5 m/s
At t=3t = 3 s: vf=2(3)+1=7v_f = 2(3) + 1 = 7 m/s

Using the work-energy theorem:
W=ΔKE=12m(vf2−vi2)=12(2)(72−52)=1(49−25)=24W = \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(2)(7^2 - 5^2) = 1(49 - 25) = 24 J

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