A small block of mass m = 0.2 kg is released from rest at a height h = 0.5 m on

Question

A small block of mass m = 0.2 kg is released from rest at a height h = 0.5 m on the frictionless curved track as shown below. At the bottom of the track is a horizontal portion, part of which is rough. The rough region is exactly L = 1.5 m long and has a coefficient of kinetic friction equal to mu_k = 0.1. At the end of the horizontal region is a spring of force constant k = 35 N/m with its right end attached to a wall. What is the maximum distance, d. by which the spring is compressed the first time that the block collides with it? How many times does the block completely cross the rough region before stopping? Show your work and indicate your reasoning. Suppose after rebounding from the spring the first time, the block reaches a height h on the curved track. which of the following actions WOULD NOT result in the block reaching a height a higher than h after rebounding from the spring the first time? Show your work and indicate your reasoning. Make the spring stiffer by increasing the spring constant, k Shorten the rough region so it is less than 1.5 m long. Express first your answers the variables m, h, mu_k, L, k, and g (acceleration due to gravity), and then their numerical values.

Explanation / Answer

As the object starts its motion on the Frictionless curved track, all the potential Energy will be converted to Kinetic energy.

mgh = 1/2 m* v2,

Mass of the object, m = 0.2 kg

Height, h = 0.5m

v = Velocity of the mass at the flat track ( Before it reaches the Surface with Friction)

mgh = 1/2 m* v2

v2 = 2gh = 2* 10m/s2 * 0.5m = 10 m2/s2

-> v = 3.16 m/s

Normal Force Acting on the mass, N = mg = 0.2* 10 = 2N

Force due to Kinetic Friction, Fk = k N = 0.1*2 = 0.2N

Deceleration due Frictional Force, a = Fk/m = 0.2/0.2 = 1m/s2

Length of the Track during which the deceleration occurs = 1.5m

Using, v2 - u2 = 2as, where v = final velocity, u = intial velocity, a = acceleration or deceleration, and s is the distance during which there is constant acceleration or deceleration.

vf2 - v2 = 2* (-1) * (1.5)

vf 2 = v2 - 3

vf 2 = 10-3 = 7m2/s2

vf = 2.645 m/s

Now, the mass goes with velocity vf and hits the spring and finally comes to a halt.

Using conservation of energy, all the kinetic energy will be converted to the potential energy of the spring.

1/2* m * vf2 = 1/2 * k * x2, where k = spring constant and x = spring compression

-> 0.2 * 7 = 35 * x2

-> x2 = 0.04

-> x = 0.2 m

The maximum distance that the spring is compressed when the block collides for the first time is 0.2m.

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