A block of mass 2 kg initially at rest moves under the action of an applied horizontal force of 6 N 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 s is (Take `g = 10 ms^(-2)`)

Options

A. 200J
B. `-200 J`
C. 600 J
D. `-600 J`

Answer

C

Explanation

The various forces acting on the block is as shown in the figure
Here, m = 2 kg, `mu`= 0.1, F = 6 N, `g = 10 ms^(-2)`
Force of friction, `f=muN=0.1 xx 2 kg xx 10 m s^(-2)=2N`
Net force with which the block moves
F'=F-f=6N - 2 N=4N
Net acceleration with which the block moves
`a=(F')/m=(4N)/(2kg)=2 m s^(-2)`
Distance travelled by the block in 10 s is
`d=1/2at^2=1/2xx2 m s^(-2)(10 s)^2 =100 m " " (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^@`
Hence, work done by the applied force,
`W_F=Fd cos theta =(6 N)(100 m) cos 0^@ =600 J`