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 $10\sec $ is (Take \[g=10~m{{s}^{-2}}\]).
$\begin{align}
  & \left( A \right)200J \\
 & \left( B \right)-200J \\
 & \left( C \right)600J \\
 & \left( D \right)-600J \\
\end{align}$

Answer 1

Hint: At first calculate the frictional force on the block and then net force so that block moves. With the help of net force solve for net acceleration with which block starts moving and then calculate distance travelled by block.

Formula used:
\[f=\mu N\]
\[d=\dfrac{1}{2}a{{t}^{2}}\]
Where:
$f$- frictional force
$\mu $- coefficient of friction
$N$- normal force on block
$a$- net acceleration of force
$t$- time period

Complete answer:

Here, values are given in question:
\[\begin{align}
  & m=2~kg,\mu =0.1, \\
 & F=6~N,g=10~m{{s}^{-2}} \\
\end{align}\]

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

Net force with which the block moves
\[\begin{align}
  & F\prime =F-f \\
 & \Rightarrow 6N-2N=4N \\
\end{align}\]

Net acceleration with which the block moves
\[a=mF\prime =\dfrac{4N}{2kg}=2~m{{s}^{-2}}\]
Distance travelled by the block in $10\sec $ is
\[\begin{align}
  & d=\dfrac{1}{2}a{{t}^{2}}=\dfrac{1}{2}\times 2~m{{s}^{-2}}{{\left( 10 \right)}^{2}} \\
 & =100~m(\therefore u=0) \\
\end{align}\]

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,
\[\begin{align}
  & {{W}_{F}}=Fdcos\theta =\left( 6N \right)\left( 100m \right)cos{{0}^{\circ }} \\
 & \implies {{W}_{F}}=600~J \\
\end{align}\]

So, the correct answer is “Option C”.

Additional Information:
Friction may be a force between two surfaces that are sliding, or trying to slip, across one another. For example, once you attempt to push a book along the ground, friction makes this difficult. Friction always works within the direction opposite to the direction during which the thing is moving, or trying to maneuver.

Note:
Calculate the frictional force first, If it's quite the applied one …. your problem becomes very easy due to frictional force, the entire system moves together. If it's not the case then the individual blocks would have individual properties, Then you'll need to consider individual FBDs then solve by considering all the forces.