Repeat the previous problem but now include friction, given a kinetic coefficient of friction of μ_k 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,
W_nc = -f_kd

This gives:

½ kX² -f_kd = mgdsin(θ)

f_k = μ_kN

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) + μ_kmgdcos(θ)

Solving for d gives:

d = kX²/{2mg [sin(θ) + μ_kcos(θ)]}

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

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