Question

A block of mass $2kg$ initially at rest moves under the action of an applied horizontal force of $6N$ on a rough horizontal surface. The coefficient of friction between block and surface is $0.1$. The work done by the applied force in $10s$ is (Take $g=10m{s}^{-2})$.

• A. $200J$
• B. $-200J$
• C. $600J$
• D. $-600J$

Answer 1

Answer: (C)

$600J$

The various forces acting on the block is as shown in the figure.
Here, $m=2kg,\mu =0.1,F=6N,g=10m{s}^{-2}$

Force of friction,
$f=\mu N=0.1\times 2kg\times 10m{s}^{-2}=2N$

Net force with which the block moves
$F^{\prime }=F-f=6N-2N=4N$

Net acceleration with which the block moves
$a=\frac{F^{\prime }}{m}=\frac{4N}{2kg}=2m{s}^{-2}$

Distance travelled by the block in $10s$ is
$d=\frac{1}{2}a{t}^2=\frac{1}{2}\times 2m{s}^{-2}(10{)}^2=100m(\therefore u=0)$

As the applied force and displacement are in the same direction therefore angle between the applied force and the displacement is $\theta =0^{\circ }$.

Hence, work done by the applied force,
$W_F=Fd\cos \theta =\left(6N\right)\left(100m\right)\cos 0^{\circ }=600J$.