Question

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

• A. $475\text{joule}$
• B. $450\text{joule}$
• C. $275\text{joule}$
• D. $250\text{joule}$

Answer 1

The correct option is A $475\text{joule}$

We know that,
work done on the object by force = change in kinetic energy

Here, work done =$\int Fdx={\int }_{20}^{30}-0.1xdx$
$=-0.1{\left[\frac{x^2}{2}\right]}_{20}^{30}=-\frac{0.1}{2}[{30}^2-{20}^2]$
$=-0.05[900-400]=-0.05\times 500=-25\text{joule}$
Applying the work energy theorem,
Work done = final kinetic energy-initial kinetic energy
Final kinetic energy= work done + initial kinetic energy
Final kinetic energy $=-25+\frac{1}{2}m{v}^2$
$=-25+\frac{10\times {10}^2}{2}$
$=475\text{J}$