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 $F=-0.1\times J/m$ during its travel from $x=20m$ to $x=30m$. Its final kinetic energy will be :

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

Answer 1

Answer: (D)

Apply work-energy theorem.
When a force acts upon a moving body, then the kinetic energy of the body increases and the increase is equal to the work done. This is work energy theorem.
Work done $=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=K_f-K_i$
Another definition of work done is force $x$ displacement.
$\therefore Fdx=K_f-\frac{1}{2}m{v}_i^2$
where the subscripts $f$ and $i$ stand for final and initial.
$F\cdot dx=K_f-\frac{1}{2}\times 10\times (10{)}^2$
$\Rightarrow F\cdot dx=K_f-500$
$\Rightarrow \underset{x=20}{\overset{x=30}{\int }}(-0.1)xdx=K_f-500$
Using the formula $\int x^{n}dx=\frac{x^{n+1}}{n+1}$, we have
$-0.1{\left[\frac{x^2}{2}\right]}_{x=20}^{x=30}=K_f-500$
$-0.1\left[\frac{(30{)}^2}{2}-\frac{(20{)}^2}{2}\right]=K_f-500$
$\Rightarrow K_f-500=-25$
$\Rightarrow K_f=500-25=475J$