Question

A block of mass $5\text{kg}$ is moving in x-direction with a constant speed of $10\text{m/sec}$. It is subjected to a retarding force $F=-0.2x\text{joule /metre}$ during its travel from $x=20\text{metre}$ to $x=30\text{metre}$. Its final kinetic energy will be

• A. $200\text{joule}$
• B. $475\text{joule}$
• C. $375\text{joule}$
• D. $400\text{joule}$

Answer 1

The correct option is A $200\text{joule}$

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

Here, work done =$\int Fdx={\int }_{20}^{30}-0.2xdx$
$=-0.2{\left[\frac{x^2}{2}\right]}_{20}^{30}=-\frac{0.2}{2}[{30}^2-{20}^2]$
$=-0.1[900-400]=-0.1\times 500=-50\text{joule}$
Applying the work energy theorem,
Work done = final kinetic energy-initial kinetic energy
Final kinetic energy= work done + initial kinetic energy
Final kinetic energy $=-50+\frac{1}{2}m{v}^2$
$=-50+\frac{5\times {10}^2}{2}$
$=200\text{J}$