Verified answer

Given :-

Mass of the body, m = 0.3 kg

Length of the plane, l = 10 m

Height of the plane, h = 5 m

Coefficient of friction, μ = 0.15

Relations :-

• N = mgcosθ

• F = mgsinθ + f [f = μN = μmgcosθ]

• F = mgsinθ + μmgcosθ

• As, length is 10 m, the value of sinθ = = and the value of cosθ = as, sin 30° = and cos 30° =

Solution :-

i. Work done by the applied force over the upward journey, = F×d

• = (mgsinθ + μmgcosθ)10

• = 10mg(sinθ + μcosθ)

• = 10×0.3×9.8 [½+ 0.15()]

• = 29.4 [ 0.5 + 0.129 ]

• = 29.4 [ 0.62 ]

• = 18.5 J

ii. Work done by gravitational force over the round trip, = -mg∆h

• = -(0.3)(9.8)(0)

• = 0

The displacement over the round trip is zero, the object ended it's journey from where it started so, work done by the gravitational force over the round trip is zero.

iii. Work done by the frictional force over the round trip, = fscos(180°)

• = (μmgcosθ)(20)(-1)

• = (0.15)(0.3)(9.8)()(20)(-1)

• = -(0.441)(0.86)(20)

• = -(0.38)(20)

• = -7.6 J

Gravitational force is conservative force, as in the solution ii. I mentioned about change in height, which states that it is not path dependent.