A disc of mass `m=50g` slides with the zero initial velocity down an inclined plane set at an angle `alpha=30^@` to the horizontal, having traversed the distance `l=50cm` along the horizontal plane, the disc stops. Find the work performed by the friction forces over the whole distance, assuming the friction coefficient `k=0.15` for both inclined and horizontal planes.

Explanation

Let s be the distance covered by the disc along the incline, from the Eq. of increment of M.E. of the disc in the field of gravity: `DeltaT+DeltaU=A_(f r)`
`0+(-mgssinalpha)=-kmgcosalphas-kmgl`
or, `s=(kl)/(sin alpha-k cos alpha)` (1)
Hence the sought work
`A_(f r)=-kmg[scosalpha+l]`
`A_(f r)=-(klmg)/(1-kcotalpha)` [Using the Eqn. (1)]
On putting the values `A_(f r)=-0.05J`