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) + μkcos(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.

Where does the block reach its maximum speed?

1. At the point (the same physical spot on the ramp) where it reached maximum speed when there was no friction. (5/33) (15%)
2. Before that point. (27/33) (82%)
3. After that point. (1/33) (3%)