The block shown in (Figure 1) has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle θ = 22.0 ∘ to the horizontal. If the block starts from rest 15.2 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?

Answer 1

h = 15.2*sin22 = 5.69 m.

V^2=Vo^2 + 2g*h = 0 + 19.6*5.69 = 111.6
V = 10.6 m/s.

Answer 2

V =106m/s