A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of `30^(@)` by a force of 10 N 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 teh inclined calculate

(a) work done against gravity
(b) work done against force of friction
(c ) increases in potential energy
(d) increases in kinetic energy
(e) work done by applied force

Explanation

m=1 kg , `theta = 30^(@) , F = 10 N , mu = 0.1 , d = 10 m `
(a) Work done against gravity ,
`W_(1)= mg xx ` Vertical distance travelled
` = mg xx d (sin theta) == (mgd) sin theta `
` = 1xx10 xx10 sin 30^(@) = 50 J `
`( :. g le 10 m//s^(2))`
(b) Work done against friction ,
`W_(2) = fxx d = muN xx s = mu mg cos theta xx d`
` = 0.1 xx1xx 10 xx cos 30^(@) xx10`
`= 10 xx 0.866 = 8.66 J `
Increase in PE = mgh = mg `(d sin theta)`
`DeltaU = 1 xx10 xx10 xxsin 30^(@)`
` = 100 xx1/2 = 50 J `
(e) Work done by applied force ,
W = Fd
= (10) (10) = 100 J
(d) By work -energy theorem , we know that work done by all the forces = change in KE
(W) = `DeltaK `
`DeltaK = W - W_(2) - DeltaU `
` = 100 - 8.66 -50 `
` = 41.34 J `