Question

A block of mass 10 kg, moving in x direction with a constant speed of 10 $m{s}^{-1}$ , is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m. It's final KE will be :

• A. 250 J
• B. 475 J
• C. 450 J
• D. 275 J

Answer 1

The correct option is B

475 J

The block of mass M = 10 kg is moving in the x - direction with a speed v = 10 m/s.

Its initial kinetic energy is

$K{E}_i=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10\times (10{)}^2=500J.$

It is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m.

Work done is given by

$W=-{\int }_{x-20}^{x-30}\overrightarrow{F}.\overrightarrow{dx}={\int }_{x-20}^{x-30}\left(0.1x\right)dx=-0.1{\left[\frac{x^2}{2}\right]}_{x-20}^{x-30}=-0.1[\frac{900}{2}-\frac{400}{2}]=0.1\times \frac{500}{2}=-0.1\times 250=-25J$

Final kinetic energy is, $K{E}_f=K{E}_i+W=500-25=475J$