# Kinetic Energy and Work: Ramp With Friction

## Homework Statement

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.15 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.22 m and the angle θ = 32.70°. Calculate the coefficient of kinetic friction for the mass on the surface. ## The Attempt at a Solution

Not quite sure how to start on this one. Any help will be appreciated. Thanks.

Sincerely,
Captformal

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First, find the energy of the mass in its initial position.

$$E = K+U = \frac{1}{2}mv_{0}^{2} + mgS_{1}\sin(\theta)$$

Now note that the block has lost all of its energy once it reaches S_{2}.

$$E + W_{other} = 0 \iff W_{other} = -E$$

The work done on the block has to be described in 2 parts: one along the path S_{1} and another along the path S_{2}. If you draw a free body diagram, you'll find that:

$$W_{1} = [mg\cos(\theta)-u_{k}mg\sin(\theta) ]S_{1}$$

$$W_{2} = -u_{k}mgS_{2}$$

After going through an algebraic mess, you'll find:

$$u_{k} = \frac{ E + mg\cos(\theta)S_{1} }{mgS_{2} + mg\sin(\theta)S_{1} }$$

Hopefully I didn't make a mistake in there somewhere.

Last edited:
Thanks Vykan12. That helped out a lot.