How Do You Calculate the Coefficient of Kinetic Friction on an Inclined Slide?

[jared69sib]

Homework Statement

A child slides down a slide with a 34 degree incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

Homework Equations

\SigmaF = ma
\SigmaF_R=\muF_N

The Attempt at a Solution

set x-axis along incline.
Got F_N=mgcos\theta and gsin\theta-gtan\thetacos\theta=a
Now I don't know what to do. I have a = 4.92 m/s², but I feel I can't proceed without \Deltax or t...

Thanks in advance!

 

[ehild]

Check your equation for a. Now it is equivalent to a=0.

Try to apply the Work-Energy theorem.

ehild

 

[HallsofIvy]

What I would do is use "energy". Taking the bottom of the slide as the 0 point for potential energy, at the top of the slide the child's gravitational potential energy is mgh where m is the child's mass and h is the vertical height of the slide. At the bottom the child's gravitational potential energy is 0 so all of that potential energy is converted to kinetic energy or lost to friction. But we are told that the speed was only 1/2 what it would be without friction. Since kinetic energy increases as the square of the speed, and the child's speed is half what it would be without friction, its kinetic energy is 1/4. If there were no friction, the kinetic energy would be equal to the original potential energy, mgh, but now is 1/4 that- mgh/4. That means that the energy lost to friction is 3mgh/4.

Let the coefficient of kinetic friction by \mu. Now you can calculate the component of the child's weight normal to the slide from the given angle and the length of the slide assuming height h and the given angle. The work done by the friction force is the coefficent of friction times the friction force times the length of the slide and that must be equal to 3mgh/4.

You will find that m, g, and h all cancel.