A body of mass 2 kg is moving along $x$-direction such that its displacement as function of time is

Question

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

Professor PC Thomas Classes · Answer

The displacement is $x(t) = t^2 + t + 1$. The velocity is the derivative of displacement: $v(t) = \frac{dx}{dt} = 2t + 1$. At $t=2	ext{ s}$, velocity is $v_1 = 2(2)+1 = 5	ext{ m/s}$. At $t=3	ext{ s}$, velocity is $v_2 = 2(3)+1 = 7	ext{ m/s}$. According to the work-energy theorem, Work done $= \Delta K = \frac{1}{2}m(v_2^2 - v_1^2) = \frac{1}{2}(2)(7^2 - 5^2) = 49 - 25 = 24	ext{ J}$.