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.
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?
im sorry it should be cos theta for all of them
and the "i" looking thing is the coefficient of friction
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
5.96 = mu*normal force
5.96 = mu * (25 cos slope)
cos slope = 4.4/5.62
mu = 5.96 *5.62/(4.4*25) = .238
Ohh I see! Thanks a lot!
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