Asked by G
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?
Can someone help me do this?
Answers
Answered by
drwls
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
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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