Question

A small block, with a mass of 250 g, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, around the loop, and then onto the final level section on the right. The maximum height of the incline is 80 cm, and the radius of the loop is 15 cm.

a.) Find the initial energy of the block.

b.) Find the velocity of the block at the bottom of the loop.

c.) Find the velocity of the block at the top of the loop.

Answer 1

(a) The initial energy of the block due to its position is 1.96 J.

(b) The velocity of the block at the bottom of the loop is 3.96 m/s.

(c)  the velocity of the block at the top of the loop is 3.13 m/s.

Initial energy of the block

The initial energy of the block due to its position is calculated as follows;

P.E = mgh

P.E = 0.25 X 9.8 X 0.8

P.E = 1.96 J

Conversation of the energy

The velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;

P.Ei + P.Ef = K.Ei + K.Ef

1.96 + 0 = 0 + ¹/₂mvf²

vf² = 2(1.96)/m

vf² = (2 x 1.96) / (0.25)

vf² = 15.68

vf = √15.68

vf = 3.96 m/s

Velocity of the block at top of the loop

The velocity of the block at the top is calculated by applying principle of conservation of energy,

P.Ei + P.Ef = K.Ei + K.Ef

1.96 = mghf + ¹/₂mvf²

where;

• hf is the position of the ball at the top of the loop = 2r = 2 x 15 cm = 30 cm = 0.3

1.96 = 0.25 x 9.8 x 0.3   +   0.5 x 0.25vf²

1.225 = 0.125vf²

vf² = 1.225/0.125

vf² = 9.8

vf = 3.13 m/s

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