JEE Main 2023 — Work Power Energy Question with Solution

The formula to calculate the change on momentum of the object is given by

$p_f-p_i=\int Fdt ...\left(1\right)$

Substitute the values of the known parameters into equation (1) and solve to calculate the final momentum of the object.

$\begin{array}{rcl}{p}_f-10 \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}& =& 2 \mathrm{N}\times 5 \mathrm{s}\\ & \Rightarrow & p_f=\left(10+10\right) \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}\\ & =& 20 \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}\end{array}$

The formula to calculate increase in kinetic energy $\left(∆K\right)$ of the particle is given by

$∆K=\frac{1}{2m}\left({p_f}^2-{p_i}^2\right) ...\left(2\right)$

Substitute the values of the known parameters into equation (2) to calculate the required increase in kinetic energy.

$\begin{array}{rcl}∆K& =& \frac{1}{2\times 5 \mathrm{kg}}\times \left({20}^2-{10}^2\right) {\mathrm{kg}}^2 {\mathrm{m}}^2 {\mathrm{s}}^{-2}\\ & =& 30 \mathrm{J}\end{array}$