# Physics

A box, with a weight of mg = 25 N, is placed at the top of a ramp and released from rest. The ramp measures 4.40 meters horizontally and 3.50 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the bottom of the ramp of 54.0 J. There is a force of kinetic friction acting on the box as it slides down the incline.

a) What is the coefficient of kinetic friction between the box and ramp? & please show work.

1. Equate the work done against friction to the mechanical (potential + kinetic) energy loss. That will allow you to solve for the friction coefficient, Uk.

The slope of the ramp is arctan 3.5/4.4 , which is 38.5 degrees.

The hypotenuse of the ramp is 5.622 m

Show your work for further assistance.

posted by drwls
2. I know the work done is

W = ì X mgcosè X d

and W is Ui+ W= Kf which is 54-(25*3.3)=-33.5

so W = ì X mgcosè X d

and so ì= (W/(mgcosè X d))

and mg=25N so what is cosè? and i think d is the hypotenuse?

posted by Jeanne
3. im sorry it should be cos theta for all of them

posted by Jeanne
4. and the "i" looking thing is the coefficient of friction

posted by Jeanne
5. It would be easier to do it the way Dr WLS suggested.
vertical distance down = 3.5 m
so potential energy lost = mgh=25*3.5 =87.5 Joules
It has Ke of 54 Joules at the bottom so lost (87.5-54) = 33.5 Joules to friction
Friction force * length of ramp = 33.5
length of ramp = sqrt(3.5^2+4.4^2) = 5.62 meters
so friction force = 33.5/5.62 = 5.96 Newtons
so
5.96 = mu*normal force
5.96 = mu * (25 cos slope)
cos slope = 4.4/5.62
so
mu = 5.96 *5.62/(4.4*25) = .238

posted by Damon
6. Ohh I see! Thanks a lot!

posted by Jeanne

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