Question bank/JEE Main/Physics

A force F= 4t acts on a particle of mass 1kg. If the particle
starts from rest the work done by the force during the first 2 second
will be 8n J, find n.

A4B6C5D3

Show answer and solution▾

$\frac{dv}{dt} = 4t$ $\int_{0}^{v}{dv} = 4\int_{0}^{t}{t\ dt}$
$V = 2t^{2}$
$ds = 2t^{2}dt$
$W = \int_{}^{}{Fds\ cos0^{\circ}}$
$= \int_{0}^{2}\mspace{2mu} 4t \cdot 2t^{2}dt$
$= 8\left\lbrack \frac{t^{4}}{4} \right\rbrack_{0}^{2}$ $= 32J$

Source: Grass-G / P-23 Physcis Paper 31 Dec FST.docx