A body of mass 10 kg is pushed up 50 cm from the ground , along a plane inclined at `45^@` to the horizontal. If coefficient of friction is 0.2 , then calculate the work done.

Explanation

Here,h=50 cm =0.5m, m=10 kg,`g=9.8m*s^(-2), theta =45^@, mu =0.2`
Let the friction acting on the body be f [Fig.1.30].

Then ,`f=mu mg cos theta`
The force against which the body is pushed up is
`F=f+mg sin theta =mg (mu cos theta + sin theta)`
The body is pushed up by a distance `h/(sin theta)` along the inclined plane.
Therefore ,the work done is
`W=(Fh)/(sin theta)=(mgh)/(sin theta)(mu cos theta +sin theta)`
`=(10xx9.8xx0.5)/(sin 45^@)xx(0.2xx cos 45^@+sin theta)`
`49xxsqrt(2) xx(0.2xx1/sqrt(2)+1/sqrt(2))=58.8 J.`