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A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by x(t) = $$\alpha t^{2} +\beta t +ym$$, where $$\alpha=1m/s^{2}, \beta=1m/s$$ and y=1m. The work done on the body during the time interval t = 2 s to t = 3 s, is _________ J.
$$x(t) = t^2 + t + 1$$ m, mass = 2 kg. Find work done from $$t = 2$$ to $$t = 3$$.
$$v(t) = \frac{dx}{dt} = 2t + 1$$ m/s
$$v(2) = 5$$ m/s, $$v(3) = 7$$ m/s
Work = Change in KE = $$\frac{1}{2}m(v_3^2 - v_2^2) = \frac{1}{2}(2)(49 - 25) = 24$$ J
The answer is Option 2: 24 J.
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