In a truck-loading station at a post office, a small 0.200-kg package is release

Question

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 4.70 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.

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

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

Explanation / Answer

a) For the motion on the horizontal surface :
initial velocity, u = 4.7 m/s
final velocity, v = 0 m/s
distance, s = 3 m
acceleration, a = ?

Use v^2 - u^2 = 2as
a = 3.68 m/s^2

Force of friction, f = uN
where u is the coefficient of friction and N is the normal reaction
Since N = mg
f = umg
Also since friction is the only force acting on the particle,
f = ma
So ma = umg

or u = a/g = 3.68/9.8 = 0.375680272

b) As the particle comes down from A to B, there is a loss in potential energy. A part of this energy is converted to kinetic energy and the remaining is used in doing work against friction.

PE = KE + Work against friction
Work against friction = PE - KE = mgh - 1/2 mv^2 = 0.927 J
(h is the radius of the circle = 1.6 m)

So the net work done by friction = Work done against friction = 0.927 J

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.