Question

A block of mass $10kg$ is moving in x-direction with a constant speed of $10m/s$. It is subjected to a retarding force $F=0.1xJ/m$ during its travel from $x=20m$ to $x=30m.$ Its final kinetic energy will be :

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

Answer 1

The correct option is A $475J$

$\text{Step 1: Applying work energy theorem}$

We can write

$W=(KE{)}_f-(KE{)}_i$

$\Rightarrow {\int }_{20}^{30}F.dx=(KE{)}_f-\frac{1}{2}m{v}^2$

$\Rightarrow (KE{)}_f=\frac{1}{2}m{v}^2+{\int }_{20}^{30}0.1xdx$

$=\frac{1}{2}m{v}^2+0.1{\left[\frac{x^2}{2}\right]}_{20}^{30}$

$\text{Step 2: Calculations}$

Putting given values

$(KE{)}_f=\frac{1}{2}\times 10\times (10{)}^2+0.1\times \left[\frac{(30{)}^2-(20{)}^2}{2}\right]$

$(KE{)}_f=475Joule$

Hence $A$ is correct option.

$\text{ALTERNATE SOLUTION:}$

$\text{Step 1: Applying Newton's 2nd law}$

$F=0.1x=ma$

$\Rightarrow a=\frac{0.1x}{m}=v\frac{dv}{dX}$

$\Rightarrow {\int }_{10}^{v}vdv={\int }_{20}^{30}\frac{0.1x}{m}dx$

$\Rightarrow [\frac{v^2}{2}-\frac{100}{2}]={\left[\frac{0.1x^2}{2m}\right]}_{20}^{30}$

$\Rightarrow \frac{v^2}{2}=50+\frac{0.1}{2m}\left(\right(30{)}^2-\left(20{)}^2\right)$

$\Rightarrow v^2=95$

$\Rightarrow (KE{)}_f=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10\times 95$

$(KE{)}_f=475J$

Hence $A$ is correct option.