Posted by **G** on Friday, March 23, 2012 at 10:34pm.

A 12 kg object is at the top of a 41 degree inclined ramp. It is released and it reaches the bottom of the 85 m ramp in 2.3 seconds. what is the force of friction between the object and the ramp? What is the coefficient of friction?

Can someone help me do this?

- Physics-11 -
**drwls**, Friday, March 23, 2012 at 11:53pm
Here's a quick way to do it using energy. The average velocity during the descent is 85/2.3 = 36.96 m/s. The final velocity is twice that, or

Vf = 73.91 m/s.

Final kinetic energy at the bottom = (M/2)Vf^2 = 32778 J

Potential energy loss at the bottom =

M*g*Lsin41 = 35859 J

Energy lost to friction = 3081 J

Since Work = Force x Distance, the friction force is

Ff = 3081J/85m = 36.2 N

That friction force, the weight and ramp angle can be used to get the friction coefficient, Mu.

Ff = M g cos41 *Mu

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