Question

problem reference 6.3
a 6.00 - kg box slides from rest down a frictionless incline from a height of 5.00 m. a constant frictional force, introduced at point a, brings the block to rest at point b, 19.0 m to the right of point a.
what is the coefficient of kinetic friction between the box and the floor. use work and energy methods to solve
0.263
0.189
0.459
0.687

Explanation:

Step1: Calculate the gravitational potential energy at the initial height

The gravitational potential energy \(U = mgh\), where \(m = 6.00\space kg\), \(g=9.8\space m/s^{2}\), and \(h = 5.00\space m\).

$$U=6\times9.8\times5=294\space J$$

Step2: Calculate the work done by friction

The work - energy theorem states that the net work done on an object is equal to the change in its kinetic energy. Since the box starts from rest and ends at rest, \(\Delta K = 0\). The work done by gravity \(W_{g}\) (which is equal to the initial potential energy \(U\)) is counteracted by the work done by friction \(W_{f}\).
The work done by friction \(W_{f}=-\mu_{k}mgd\), where \(d = 19.0\space m\)

Since \(W_{g}+W_{f}=\Delta K = 0\), we have \(mgh-\mu_{k}mgd = 0\)

Cancel out \(mg\) from both sides of the equation: \(h-\mu_{k}d=0\)

Then \(\mu_{k}=\frac{h}{d}\)

Substitute \(h = 5\space m\) and \(d = 19\space m\)
\(\mu_{k}=\frac{5}{19}\approx0.263\)

Answer:

0.263