Posted by **Jeanne** on Sunday, October 30, 2011 at 4:02pm.

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.

- Physics -
**drwls**, Sunday, October 30, 2011 at 4:09pm
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.

- Physics -
**Jeanne**, Sunday, October 30, 2011 at 4:14pm
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?

- Physics -
**Jeanne**, Sunday, October 30, 2011 at 4:15pm
im sorry it should be cos theta for all of them

- Physics -
**Jeanne**, Sunday, October 30, 2011 at 4:15pm
and the "i" looking thing is the coefficient of friction

- Physics -
**Damon**, Sunday, October 30, 2011 at 4:36pm
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

- Physics -
**Jeanne**, Sunday, October 30, 2011 at 4:50pm
Ohh I see! Thanks a lot!

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