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 (87.554) = 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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