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Question
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 the incline, calulate
- work done against gravity
- work done against force of friction
- increase in potential energy
- increase in kinetic energy
- work done by applied force.
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Solution
Consider the adjacent diagram the block is pushed up by applying a force F.
Normal reaction (N) and frictional force (f) is shown.
Given, mass = m = 1 kg, θ = 30°
F = 10 N, μ = 0.1 and s = distance moved by the block along the inclined plane = 10 m
a. Work done against gravity = Increase in PE of the block
= mg × Vertical distance travelled
= mg × s(sin θ) = (mgs) sin θ
= 1 × 10 × 10 × sin 30° = 50 j ....(∵ g ≤ 10 m/s2)
b. Work done against friction
wf = f × s = μN × s = μ mg cos θ × s
= 0.1 × 1 × 10 × cos 30° × 10
= 10 × 0.866
= 8.66 J
c. Increase in PE = mgh = mg (s sin θ)
= 1 × 10 × 10 × sin 30°
= `100 xx 1/2`
= 50 J
d. By the work-energy theorem, we know that work done by all the forces = change in KE
(W) = ΔK
ΔK = Wg + Wf + Wf
⇒ = – mgh – fs + FS
= – 50 – 8.66 + 10 × 10
= – 50 – 8.66
= 41.34 J
e. Work done by the applied force, F = FS
= (10)(10)
= 100 J
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