A block weighing 250N is pulled over a horizontal plane at a constant velocity up to a distance of 10 m. The coefficient of kinetic friction is 0.2 and the force is applied by a string, attached with the block, inclined at `60^@` with the vertical . Find the work done against friction.

Explanation

Let the force applied on the block be F [Fig.1.24].
Horizontal component of the applied force along the plane `=F sin 60^@ =sqrt(3)/2 F ` and its vertical component `= F cos 60^@ =F/2`.

Since there is no vertical acceleration of the block, net force acting vertically is zero.
i.e., `R+F cos 60^@ =W`
[where R is the normal reaction on the block]
`therefore R=W =F/2`
As the body is moving with a uniform velocity, the horizontal component of applied force= frictional force
`or, sqrt(3) /2F= mu R =mu(W-F cos 60^@)=0.2 (250-F/2) =50-0.1 F`
`therefore F=(50)/(0.866+0.1)=(50)/(0.966) N`
Hence , work done by the applied force
`=F sin 60^@ xx10 =(50)/(0.966)xxsqrt(3)/2xx10=448.25 J`.