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**1. A 2 kg object slides down a frictionless track (starting 0.27 m above the ground) to a horizontal surface where it collides elastically with a 0.5 kg mass. 3.3 m later the 0.5 kg mass hits a k=25 N/m spring and compresses is 6 cm. What is the coefficient of friction on the horizontal surface?**

**2. GPE=mgh, KE=1/2*mv^2, p=mv, Momentum and Kinetic Energy are conserved in this problem, F=kx, and W=1/2*kx^2**

**3. First I found out that the velocity of the box right before collision was 2.300434742 m/s, because of conservation of energy**

Then I figured out the velocity of the 0.5 kg box to be 3.680695766 m/s from 2 equations and 2 variables and using the fact that it is a 1 D Elastic Collision.

Then I found out that mu=-a/g, and so I learned that the energy done by friction was 1/2*-9.8*mu*3.3=-16.17 mu, right before it hits spring

Then, I found out that the work done by the box when it hits the spring is 1/2*25*.06^2=0.045J, so the energy left right before it hits the spring is 0.045J, and because the energy right after collision is 1/2*0.5*3.680695766^2, we see that -16.17mu=0.045-3.38688033, and so mu=.2066716345

Then I figured out the velocity of the 0.5 kg box to be 3.680695766 m/s from 2 equations and 2 variables and using the fact that it is a 1 D Elastic Collision.

Then I found out that mu=-a/g, and so I learned that the energy done by friction was 1/2*-9.8*mu*3.3=-16.17 mu, right before it hits spring

Then, I found out that the work done by the box when it hits the spring is 1/2*25*.06^2=0.045J, so the energy left right before it hits the spring is 0.045J, and because the energy right after collision is 1/2*0.5*3.680695766^2, we see that -16.17mu=0.045-3.38688033, and so mu=.2066716345

However, this is wrong, so I was wondering if ny assumption that the work remaining is 0.045 right before the contact with spring was wrong.

Thanks