Q. A block of mass $10kg$ is moving with a speed of $10\frac{ m}{s}$ in $x-$ direction. A retarding force $F=-0.1x\frac{J}{m}$ acts on it from $x=20m$ to $x=30m$ . Find its final kinetic energy.
NTA AbhyasNTA Abhyas 2020
Solution:
Work done by variable force is $W_{F}=\displaystyle \int _{x_{1}}^{x_{2}}Fdx=-0.1\displaystyle \int _{20}^{30}xdx$ $W_{F}=-\frac{0 . 1}{2}\left[x^{2}\right]_{20}^{30}=-25J$
By work energy theorem
$W_{F }=\Delta KE$ $-25=KE-\frac{1}{2}\times 10\times \left(\right.10\left(\left.\right)^{2}$
So, $KE=475J$
$W_{F }=\Delta KE$ $-25=KE-\frac{1}{2}\times 10\times \left(\right.10\left(\left.\right)^{2}$
So, $KE=475J$