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

1. work done against gravity
2. work done against force of friction
3. increase in potential energy
4. increase in kinetic energy
5. work done by applied force.

Answer 1

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 = W_g + W_f + W_f

⇒ = – 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