Question

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

• A. 475 joule
• B. 450 joule
• C. 275 jolule
• D. 250 jolule

Answer 1

The correct option is A 475 joule

We know that,
Change in kinetic energy
= work done on the object by force
Here, work done = $\int Fdx={\int }_{20}^{30}-0.1\text{x dx}$
$=-0.1{\left[\frac{x^2}{2}\right]}_{20}^{30}=-\frac{0.1}{2}[{30}^2-{20}^2]$
$=-0.5[900-400]=-0.05\times 500=-25\text{Joule}$
Now, initial kinetic energy
$=\frac{1}{2}\times 10\times 100=500\text{Joule}$
Final kinetic energy $=500-25=475\text{Joule}$