Interesting friction physics problem

Question

The following task was discussed on an internet site, and I also became curious about its solution, but I simply can't figure it out:

On a horizontal, rough table, the coefficient of sliding friction μ is a function of the distance x from the table's edge. Starting a small body from the edge with different initial velocities v, we find that the distance travelled until it stops has the form s=kv, where k is the parameter characteristic of the table. Let's determine the location dependence of the friction coefficient!

Thank you for any help!

Answer 1

This question requires calculus. Note that the force of friction is going to be F(x) = m g µ(x), where x is the distance from the edge of the table. The initial kinetic energy of the motion is 1/2 m v² and it's all horizontal. We need to find the point such that the work done by friction is the same as the kinetic energy and, hence, 1/2 m v² = ∫dx m g µ(x). Rearranging, we find that

v²/(2g) = ∫dx µ(x)

where the integral is from 0 to k v. Now, guess μ(x) = a x + b. Then,

v²/(2g) = 1/2 a (kv)² + b(kv).

choosing a = 1 / (g k²), b =0 satisfies the equality. Thus, µ(k) = x/(gk²).

*Update: Alternative approach using the Leibniz integral rule*

Go back to v²/(2g) = ∫dx µ(x), we know that the derivative of the LHS with respect to v is v/g. Doing d/dv = k d/ds on the RHS and applying the Leibniz rule we get that d/dv ∫dx m g µ(x) = k µ(s). Substitute v/g = s/(kg) into the LHS and equate both sides to end up with the expression from above.