What is the Coefficient of Kinetic Friction on a Horizontal Surface?

[javacola]

Homework Statement

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 ). 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 4.10 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?

Homework Equations

1/2mv^2

The Attempt at a Solution

A) No idea

B) (1/2)*(0.2kg)*(4.1m/s)^2 - (0.2kg)*(9.8m/s2)*(1.6m)

 

[Aureolux]

Here are a couple equations which you should find useful for part (a):

F_{fk} = -\mu_k F_N
(where F_{fk} is the force due to kinetic friction, \mu_k is the coefficient of kinetic friction, and F_N is the normal force of the floor on the package; the - denotes that the force of friction opposes the direction of motion)

F = ma
(Newton's second law)

v_f^2 = v_i^2 + 2a(x_f - x_i)
(relevant one-dimensional equation of motion for constant acceleration)