A block of mass 0.73kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.73kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.46. This section (from points B to C) is 3.92m in length. The block then enters a frictionless loop of radius r= 2.85m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing below is not to scale.

What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track?

What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?

A block of mass 0.73kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.46. This section (from points B to C) is 3.92m in length. The block then enters a frictionless loop of radius r= 2.85m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing below is not to scale. What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track? What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?

Explanation / Answer

The mass of block, m = 0.73 kg

Since it initially at rest, initial velocity u = 0

From the law of conservation of energy, we have

Total energy at A = Total energy at B

m g h + (1/2)mu^2 = (1/2)mv^2

v^2  = 2 g h / m = 19.6h / 0.73 = 26.85h

v = 5.18(h)^(1/2)

The velocity at point D must be equal to (gr)^(1/2) = (9.8 * 2.85)^(1/2) = 5.28 m/s

This velocity must be equal to the velocity of block at B.

So, 5.18(h)^(1/2) = 5.28

h = 1.04 m

So the minimum kinetic energy of block at B, = (1/2)mv^2 = 0.5 * 0.73 * 5.28 = 1.93 J

As the above calculation we confirm that the minimum height from which the block should

start inorder to have eneogh speed at point D that the block will not leve the track is

h = 1.04 m

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