Question

A 0.50 kg ball moves from rest without friction down a ramp and then travels around the inside of a circular loop of radius R = 15.0 cm. The initial elevation of the ball on the ramp is h. If the speed of the ball at point B (at a height of 30.0 cm from the ground) is 3.43 m/s, find h.

Answer 1

Total energy when the ball is at a height h on the ramp:

T₁ = potential energy + kinetic energy

T₁ = m g h + (1/2) m v²

Since it starts from rest , so v = 0

T₁ = m g h =0.5 x 9.81 x h

Now total energy when the ball is at point B at a height of 30.0 cm from the ground:

T₂ = potential energy + kinetic energy

T₂ = m g H + (1/2) m v² (where H = 30 cm = 0.3 m)

T₂ = 0.5 x 9.81 x 0.3 + (1/2) x 0.5 x 3.43² = 4.41 J

Now by the law of conservation of energy:

T₁ = T₂

0.5 x 9.81 x h = 4.41

h = 0.9 m = 90 cm