Question

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.

Answer 1

Retadation on horizontal surface,
`a_(1) =mu =0.15 xx 10 = 1.5 m//s^(2)`
Velocity just entering before horizontal surface,
`v=sqrt(2a_(1)S_(1))`
`=sqrt(2 xx 1.5 xx 0.5)`
`sqrt(1.5) m//s`
.
Acceleration on inclined plane,
`a_(2) =g sin theta - mug cos theta`
`=10 xx 1/2-0.15 xx 10 xx (sqrt3)/2`
`=3.7 m//s`
`v=sqrt(2a_(2)S_(2)) =sqrt1.5`
`:. S_(2)=(1.5)/(2a_(2))=(1.5)/(2xx 3.7)=0.2 m`
`h=S_(2) sin 30^@=0.1 m`
Now work done by friction
`=-[initial mechanical energy"]`
`=-mgh =-(0.05)(10)(0.1)`
`=-0.05 J`