# Energy problem (compression of a spring)

1. Nov 1, 2014

### Wellenbrecher

1. The problem statement, all variables and given/known data
A 4.29-kg ball of clay is thrown downward from a height of 2.53 m with a speed of 4.21 m/s onto a spring with k = 1770 N/m. The clay compresses the spring a certain maximum amount before momentarily stopping.

a) Find the maximum compression of the spring.
b) Find the total work done on the clay during the spring's compression.

2. Relevant equations
Ep = mgh
Ek = 1/2mv2
Espring = 1/2kx2 (k: spring constant, x: compression length)

3. The attempt at a solution
I tried two equations but I couldn't find the true answer.
First equation I solved is
mgh + 1/2mv2 = 1/2kx2
And the second one is
mg(h+x) + 1/2mv2 = 1/2kx2

3. Extra Info
Following answers are FALSE for question a:
0,1234982833 m
0,49698749 m
0,40406548 m

Following answers are FALSE for question b:
271,740459713 J
144,4929915 J
-162,5281058 J

2. Nov 1, 2014

### Staff: Mentor

Can you show some additional details about how you calculated your attempts? We need to see how you are applying the Relevant Equations.

Also, 10 digits of accuracy is a bit much when the given values have three significant figures. You might want to round your results to match the given number of significant figures.

3. Nov 1, 2014

### quantumtimeleap

Hi there!

It would have been clearer and easier for us to help you if you indicated what numerical answers you got using which equation, rather than just putting the numbers out there.

Anyway, I think you have the equations right. Just equate the energy of the clay at the top ( $E = KE + U_{grav}$ ) and the energy at the bottom ($E = U_{elastic}$). The second equation you identified in your attempt is correct, because you set $U_g = 0$ at the point of maximum compression of the spring.

Are you sure your answer is wrong? Maybe you only made a mistake with the units or sign (negative or positive) somewhere?

4. Nov 1, 2014

### stevendaryl

Staff Emeritus
Hmm. I believe your formulas are correct. The difference between mgh and mg(h+x) is only whether you take into account the extra distance the mass falls as it compresses the spring. I don't think for this problem there will be much difference in the final answers.

I think you may have just made an arithmetic or algebraic mistake, or else used the wrong value for one of the constants. Could you write out your answer symbolically, giving x in terms of m, g, h and v?

5. Nov 1, 2014

### Wellenbrecher

For mgh + 1/2mv2 = 1/2kx2
4,29*9,81*2,53 + 1/2*4,29*4,212 = 1/2*1770*x2
144,4929915 = 885*x2
x2 = 144,4929915/885
x2 = 0,1632689169491525423728813559322
x = 0,40406548596626332293541051481052 m

For mg(h+x) + 1/2mv2 = 1/2kx2
4,29*9,81*(2,53+x) + 1/2*4,29*4,212 = 1/2*1770*x2
42,0849*(2,53+x) + 38,0181945 = 885*x2
42,0849*(2,53+x) + 38,0181945 = 885*x2
42,0849*x + 144,4929915 = 885*x2
885*x2 - 42,0849*x - 144,4929915 = 0

x0 = -0.380988 x1 = 0,428541 m

I haven't found 0,428541 m before. Can someone check my solution and look for arithmetic mistakes? Since I've used 3 out of 4 tries, I can't dare to post it to the system. If you confirm this answer, I will post it.

And please teach me how to solve the question b. Is the equation for it "-(mg(h+x) + 1/2mv2) J" ?

6. Nov 1, 2014

### haruspex

I don't know what your x0 and x1 are, but the number is near enough right. (I get 0,42836...)
You are only given data to three significant figures, so you should not have more significant figures in your answer. It implies too great a degree of accuracy.
Yes.

7. Nov 1, 2014

### Wellenbrecher

System accepted 0,42854 m for question a. Thanks everyone :)