An object, previously at rest, slides #3 m# down a ramp, with an incline of #(pi)/3 #, and then slides horizontally on the floor for another #25 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Question

Julia Aguilar · Accepted Answer

0.098

All of the gravitational potential energy was transformed into heat by friction due to energy conservation. #Delta "GPE" = mg (sin(pi/3)*3"m")# #"Work done by friction" = Sigma ("Friction" * "distance")# #= mg cos(pi/3) mu(3"m") + mg mu(25"m")# Thus, comparing them yields #mg (sin(pi/3)*3"m") = mgcos(pi/3) mu(3"m") + mg mu(25"m")# #sin(pi/3) * 3"m" = mu(cos(pi/3)*(3"m") + (25"m"))# #mu = frac{sin(pi/3) * 3"m"}{cos(pi/3)*(3"m") + (25"m")}# # = frac{sqrt3/2*3"m"}{1/2*(3"m") + (25"m")}# #~~ 0.098#

Leo Abeyta · Answer

undefined The total distance traveled by the object is 28 meters. We can use the work-energy principle to find the kinetic friction coefficient. The work done by friction is equal to the change in kinetic energy.

The change in kinetic energy is equal to the work done by gravity on the ramp minus the work done by friction on the ramp.

The work done by gravity on the ramp is equal to the change in gravitational potential energy, which is given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the ramp.

The work done by friction on the ramp is equal to the frictional force multiplied by the distance traveled on the ramp.

The work done by gravity on the ramp is $mgh = mgd\sin(\frac{\pi}{3})$, where $d$ is the length of the ramp.

The work done by friction on the ramp is $F_{\text{friction}} \cdot d$, where $F_{\text{friction}}$ is the frictional force on the ramp.

The work done by friction on the floor is $F_{\text{friction}} \cdot d_{\text{floor}}$, where $d_{\text{floor}}$ is the distance traveled on the floor.

Since the material of the ramp and the floor are the same, the kinetic friction coefficients are equal. Let's call this coefficient $μ$.

We can now set up the equation:

$$mgd\sin(\frac{\pi}{3}) - F_{\text{friction}} \cdot d = \frac{1}{2}mv^2 + F_{\text{friction}} \cdot d_{\text{floor}}$$

Solving for $μ$, we get:

$$μ = \frac{mgh - \frac{1}{2}mv^2}{mgd\sin(\frac{\pi}{3}) + mgd_{\text{floor}}}$$

Substituting the given values, we can solve for $μ$.