Question

A force acts on a \(2~\text{kg}\) object so that its position is given as a function of time as \(x=3t^2+5\). What is the work done by this force in first \(5\) seconds ?
\([\text{Mains-2019, Jan 9th, Shift 2}]\)

Answer 1

Position, \(x=3t^2+5 \\ \)
\(\therefore \text{Velocity}, v=\dfrac{dx}{dt} \\ \)
\(\Rightarrow v=\dfrac{d(3t^2+5)}{dt} \\ \)
\(\Rightarrow v=6t+0 \\ \)
\(\begin{matrix}
\text{At }~ t =0 ~\text{sec} &v_i=0~\text{m/s} \\
\text{At }~ t=5 ~\text{sec}&v_f=30~\text{m/s}
\end{matrix} \\ \)
According to work-energy theorem, \(W=\Delta KE. \)
\(W=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2\\W=\dfrac{1}{2}(2)(30)^2-0=900~\text{J} \)