Question

A block of mass $1.0\, kg$ moving on a horizontal surface with speed $2 \,m / s$ enters a rough surface. The retarding force $\left(F_{r}\right)$ on the block is given by
$F_{r}=\begin{cases}-k / x ; 10\, m< x< 100\, m \\ 0 ; x ; x<10 m \text { and } x > 100 \,m\end{cases}$
where, $k=0.5\, J$. The kinetic energy of the block at $x=100\, m$ is

• A. 4.5 J
• B. 2.5 J
• C. 0.5 J
• D. 1.5 J

Answer 1

Answer: (D)

The given, $m=1.0 \,kg , u=2\, m / s$
$k=0.5 \,J , x=100 \,m , v =?$
Kinetic energy $=?$
$F_{r}=-\frac{k}{x}=m a$
$\Rightarrow -\frac{0.5}{100}=1 \times a$
$\Rightarrow a=-\frac{1}{200} \,m / s ^{2}$
$v^{2}=u^{2}+2 a x$
$=2^{2}-2 \times \frac{1}{200} \times 100$
$v^{2}=4-1$
$v^{2}=3 m^{2} / s ^{2}$
Kinetic energy $=\frac{1}{2} m v^{2}$
$=\frac{1}{2} \times 1 \times 3=1.5\, J$