# Homework Help: Spring/ Kinetic Energy Problem

Tags:
1. Oct 30, 2015

### Elvis 123456789

1. The problem statement, all variables and given/known data
At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. The spring with a force constant k = 4300 N/m and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 68.0 kg are pushed against the other end, compressing the spring 0.390 m . The sled is then released with zero initial velocity.

1.)What is the sled's speed when the spring returns to its uncompressed length?

2.)What is the sled's speed when the spring is still compressed 0.225 m ?

2. Relevant equations
Work done by a spring = 1/2*K*X^2

Work = change in Kinetic Energy

3. The attempt at a solution

1.) 1/2*K*X^2=K2 - K1 ------> K1=0

1/2*K*X^2=1/2*M*V2^2

V2= sqrt (K*X^2)/M = sqrt (4300*(0.39^2))/68 = 3.1 m/s

2.) I tried the same thing from the part before but instead i chose X=0.165m since thats the distance that the spring moves from the initial position when the spring is compressed 0.39m. I got 1.31 m/s which was wrong and I don't know why

2. Oct 30, 2015

### haruspex

That does not work. The force decreases as the spring expands, so there is relatively more energy released in the first part than in the second.
Instead, just consider the total energy at the 0.225m of compression point.

3. Oct 30, 2015

### Elvis 123456789

`
So then could I setup the problem as follows

1/2*K*X^2=1/2*M*V2^2 - 1/2*M*V1^2 ----> V1 = sqrt (M*V2^2 - K*X^2)/M

using V2 = 3.1 m/s
X = 0.2 m

V1= 2.66 m/s

is this correct?

4. Oct 30, 2015

### haruspex

That's the right idea, but a little inaccurate. You are given data to three significant figures. Keep 4 sig figures through all your working, and write the final answers to three.

Edit: just noticed you wrote x=.2 in your last post. The given value was .225.

5. Oct 31, 2015

### Elvis 123456789

oh yah, i mixed up the 0.2 for another problem I was looking at, i meant 0.255.

Do you think you can clarify a little bit more on why what i did the first time was wrong. Like why did it work for the first part and not the second.

6. Oct 31, 2015

### haruspex

Consider two different extension of a spring, x and y. In your first approach you took the difference in energies as $\frac 12k(y-x)^2$. But the energy difference is clearly $\frac 12k(y^2-x^2)$. Are they the same?

7. Oct 31, 2015

### Elvis 123456789

oh ok I see. So whenever I am to obtain the distance traveled by the spring, I take the difference of the squares with respect to the initial position?

8. Oct 31, 2015

### haruspex

For energy, yes. Not for difference in force, obviously.

9. Oct 31, 2015

### Elvis 123456789

ok I see. Thanks for the help haruspex!