A block of mass 1kg is pushed up a surface inclined to horizontal at an angle of `30^(@)` by a force of 10N parallel to the inclined surface `[figure]`. The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
(a) work done against gravity
(b) work done against force of friction
(c) increase in potential energy
(d) increase in kinetic energy
(e) work done by applied force.

Explanation

In figure, f is the force of friction acting down the incline.
Here, `m=1kg, theta=30^(@)`
`F=10N, mu=0.1, s =10m`. (a) Work done against gravity
`W_(g) = mg sin theta xxs=1xx10 sin 30^(@)xx10=(100)/(2)=50J`
(b) Work done against friction
`W_(f)=fxxs=mu Rxxs=mu mg cos theta xxs`
`=0.1xx1xx10 cos 30^(@) xx 10 =10xx0.866=8.66J`
(c) Increase in P.E. `=mgh=mg(s sin theta )=1xx10xx10 sin 30^(@) =50J`
(d) Increase in K.E. `=Delta K=?`
Acceleration of block up the incline
`a=(F-(mg sin theta +mu mg cos theta))/(m)=(10-(1xx10xx(1)/(2)+0.1xx10xx0.866))/(1)=4.13m//s^(2)`
From `upsilon^(2)-u^(2)=2as`
`DeltaK=(1)/(2)m(upsilon^(2)-u^(2))=(1)/(2)mxx2as=mas =1xx4.13xx10=41.3J`