Add friction

Repeat the previous problem but now include friction, given a kinetic coefficient of friction of μk between the block and the ramp. What is d now?






The energy conservation equation, which was:

½ kX2 = mgdsin(θ)

needs to be modified by the work done by friction,
Wnc = -fkd

This gives:

½ kX2 -fkd = mgdsin(θ)

fk = μkN

Drawing a free-body diagram of the block and summing forces in a direction perpendicular to the ramp, it can be shown that:

N = mgcos(θ)

The energy equation becomes:

½ kX2 = mgdsin(θ) + μkmgdcos(θ)

Solving for d gives:

d = kX2/{2mg [sin(θ) + μkcos(θ)]}

For a particular case where k = 50 N/m, X = 0.10 m, m = 0.080 kg, θ = 30 degrees, and μk = 0.50, d works out to:

d = 0.34 m instead of the 0.64 m we got without friction.