posted by G on .
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?
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