In a truck-loading station at a post office, a small 0.200 kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (Figure 1). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.
A). What is the coefficient of kinetic friction on the horizontal surface?
b How much work is done on the package by friction as it slides down the circular arc from A to B ?

Question

03-08-2022
Physics

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In a truck-loading station at a post office, a small 0.200 kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m (Figure 1). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.
A). What is the coefficient of kinetic friction on the horizontal surface?
b How much work is done on the package by friction as it slides down the circular arc from A to B ?

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mickymike92 mickymike92 · Answer 1 · 2022-08-03T19:24:44+02:00

The coefficient of kinetic friction on the horizontal surface, μ = 0.39.

The work done on the package by friction as it slides down the circular arc from A to B, W = -4.176.

What is the frictional force acting on the package moving on the track?

Frictional force is the force which opposes the motion of an object over another.

Frictional force acts at the surface of contact of the objects.

The frictional force, F is related to the coefficient of kinetic friction on the horizontal surface, μ, and the normal reaction or weight, mg, of the package as follows:

F = μmg

Work done against friction, W = F * d

W = μmgd

The Kinetic energy of the package is equal to the work done against friction.

Kinetic energy of the package = mv²/2

μmgd = mv²/2

μ = v²/2gd

μ = 4.8²/(2 * 9.81 * 3)

μ = 0.39

b. Work done on the package by friction as it slides down the circular arc from A to B is given as follows:

W = mv²/2 - mgh

W = v²/2 - gh

W = 4.8²/2 - 9.81 * 1.6

W = -4.176

In conclusion, the work done by friction is lost in the form of heat.

Learn more about coefficient of kinetic friction at: https://brainly.com/question/20241845

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