q7.14 In a truck-loading station at a post office, a small 0.200-kg package is r

Question

q7.14

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 (the figure (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 Bwith a speed of 4.60 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.

Part A

What is the coefficient of kinetic friction on the horizontal surface?

Part B

How much work is done on the package by friction as it slides down the circular arc from A to B?

Explanation / Answer

part A)

let the coefficient of kinetic friction is u

friction force at the surface is

f = u*mg

Now, Using work energy theorum

change in kinetic energy = work done by friction

0 -0.5 * m * 4.60^2 = - u*m * 9.8 * 3

u = 0.719

the coefficient of kinetic friction is 0.719

B)

let work done by friction is Wf

change in mechincal energy = Wf

0.5 * m*v^2 - mgh = Wf

0.5 * 0.2 * 4.6^2 - 0.2 * 9.8 * 1.60 = Wf

Wf = -1.02 J

-1.02 J work is done on the package by friction as it slides down the circular arc from A to B

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