Question

200N force was used to move 150N block on an inclined plane of length 12m with a height of 4m. The efficiency of the inclined plane will be

Answer 1

42.1%

Step-by-step explanation:

Applying,

E(%) = (M.A/V.R)×100............. Equation 1

Where E(%) = Percentage efficiency, M.A = mechanical advantage, V.R = Velocity ratio.

But,

M.A = L/E............... Equation 2

Where L = Load, E = Effort.

From the question,

Given: L = 150 N, E = 200 N

Substitute these values into equation 2

M.A = 200/150

M.A = 1.33

Also,

V.R for Inclined plane = 1/sin∅ =

V.R = 1/sin∅............... Equation 2

Where ∅ = angle of the inclined plane.

Where,

tan∅ = 4/12

∅ = tan⁻¹(4/12)

∅ = 18.43°

Therefore,

V.R = 1/sin18.43

V.R = 3.16.

Substituting the value of M.A and V.R into equation 1

E(%) = (1.33/3.16)×100

E(%) = 42.1%