A body of mass 0.3 kg is taken up an inclined plane of length 10m and height 5 m and then allowed to slide down to bottom again. Find
work done by frictional force over the round trip if `mu=0.15`.

Explanation

`sintheta=(5)/(10)=(1)/(2)=sin.(pi)/(6)` so, `theta=(pi)/(6)`
Work done by frictional force, `W=2vecf.vecl`
`W=2flcosalpha(alpha=pi,cosalpha=-1)`
`W=-2fl=-2muRl=-2mumgcosthetal`
`W=-2xx0.15xx0.3xxcos.(pi)/(6)xx10xx9.8`
`=-7.6J`