Question

A block of metal of mass 50 g when placed over an inclined plane at angle of 15º with horizontal, slides down without acceleration. Find the acceleration if the angle is increased by 15º.

Answer 1

m = 50 g = 5 x 10^-2 kg

Angle of repose, α = 15º

Now, μ = tanα = tan 15º = 0.2679

Now, the increased angle of inclination,

θ = 15º + 15º = 30º

Let a = acceleration of the block in the downward direction.

Now, net increased downward force acting on the block is

F₁ = mg sinθ - F

But, F = μR = μ mg cosθ

Hence F₁ = mg sinθ - μ mg sinθ

Therefore a = F₁/m = g sinθ - μ g cosθ = g (sinθ - μ cosθ)

= 9.8(sin 30º - 0.2679 cos 30º)

= 9.8(0.5000 - 0.2679 x 0.866)

= 9.8(0.5000 - 0.2370)

= 9.8 x 0.5320 = 52 ms^-2