Determining coefficient of friction

quasar7744

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

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

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Shooting Star

Homework Helper
No, that part is quite correct.

The correct eqn is:

KE before friction - energy lost due to friction = 0.045 =>
3.386 -16.17mu = 0.045 =>
mu = 0.212.

(You need not retain so many decimal places!)

quasar7744

No, that part is quite correct.

The correct eqn is:

KE before friction - energy lost due to friction = 0.045 =>
3.386 -16.17mu = 0.045 =>
mu = 0.212.

(You need not retain so many decimal places!)

I don't understand how you got 0.212 as the answer, and when I solve that equation, I still get the wrong answer.

Thanks

Shooting Star

Homework Helper
I don't understand how you got 0.212 as the answer, and when I solve that equation, I still get the wrong answer.

Thanks
I got the answer by solving the eqn I've written down. Do you have some doubts about that eqn? Tell me.

Also, I didn't understand, you still get the wrong answer, when you solve which eqn?

BTW, do you know the correct ans?

quasar7744

I got the answer by solving the eqn I've written down. Do you have some doubts about that eqn? Tell me.

Also, I didn't understand, you still get the wrong answer, when you solve which eqn?

BTW, do you know the correct ans?

the equation I was referring to was 3.386-16.17mu=0.045, when I solve it, I get 0.206,

And the correct answer is .2050 to the nearest ten thousandths.

Last edited:

Shooting Star

Homework Helper
Yes, I had made a mistake in +/- and got the wrong result of 0.212. The ans according to what I've given is 0.206, the same as your value. Sorry. Actually, we have both written down the same eqn.

Was there any stipulation in the problem to consider all quantities up to a certain number of decimal places? I've never seen any problem where you have to find out speeds up to 9 decimal places. Then other things should be more accurate, e.g., g=9.81 m/s/s etc.

quasar7744

Yes, I had made a mistake in +/- and got the wrong result of 0.212. The ans according to what I've given is 0.206, the same as your value. Sorry. Actually, we have both written down the same eqn.

Was there any stipulation in the problem to consider all quantities up to a certain number of decimal places? I've never seen any problem where you have to find out speeds up to 9 decimal places. Then other things should be more accurate, e.g., g=9.81 m/s/s etc.
Perhaps we need to consider the work of friction even when it is in contact with the spring.

But I don't know how to do this part. And no there was no stipulation. I am pretty sure that the velocities are all correct.

Shooting Star

Homework Helper
You can try it. Then energy before impact with spring = energy to compress spring by 6 cm + energy lost due to friction in moving 6 cm = 0.5*25*0.06^2 + mu*m*g*0.06.

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