Work Energy Friction

1. Dec 21, 2005

joex444

Just need a simple logic check here. Let's say a block of 2kg is positioned on an incline of 53 degrees. 4m from the block is a spring with a k=70N/m. the coefficient of kinetic friction is 0.36. How fast is the block moving when it hits the spring? How far is the spring compressed.

Ok, so find the height from the block to the spring (this is <4m), and say Ugrav = Wfriction + KE, to answer the first part. Then, that KE = Uspring + Wfriction to find the compression. Now, Wfriction would be Ff(x), where x is the compression of the spring. Yes? Or, would KE = Uspring be the correct way?

2. Dec 21, 2005

Staff: Mentor

Sounds good.

When solving for the compression, include the work done against friction. But don't forget the change in gravitational PE.

3. Dec 21, 2005

andrewchang

you may want to try solving the second part of this problem by skipping the kinetic energy- going directly from its original position to the final position. So, if x were the distance that the spring were compressed:

$$U + W_f = EPE$$
$$mg(h+x) + W_f = 0.5kx^2$$

4. Dec 22, 2005

joex444

Hmm...forgot about gravity after it hits the spring. Ok, so (H and X are the distance on the inclined plane, so I really want Hx and Xx in Ugrav)...
$$mg(h+x)\sin\theta + \mu mg\sin\theta = \frac{1}{2} kx^2$$
and I know m, g, h, theta, mu, and k, so 1 variable left x. I ended up with a quadratic, once I put numbers into it: $$0=35x^2-15.65x-21.28$$ so x=1.03m. The question did say it was a "long" spring. I forgot gravity on the 2nd part on the final, so I ended up with an answer of 0.802m...

5. Dec 22, 2005

mukundpa

Wait; check the work by(or against) friction. The term does not appear dimensionally correct.

6. Dec 22, 2005

Staff: Mentor

Several problems with the 2nd term (the work done by friction):
(1) As mukundpa points out, it is dimensionally incorrect. (You forgot the distance.)
(2) The sign is incorrect.
(3) The sine is incorrect.