# Spring energy to kinetic energy

1. Mar 1, 2012

### shrutij

1. The problem statement, all variables and given/known data
A horizontal spring with spring constant 92.2 N/m is compressed 15.5 cm and used to launch a 2.93 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.121. How far does the box slide across the rough surface before stopping?

2. Relevant equations
KE=1/2mv^2
Usp=1/2kx^2
Work-energy theorem

3. The attempt at a solution
The Kinetic energy gained by the box should be equal to the potential energy stored in the spring, so 1/2kx^2=1/2mv^2, substituting all values in, I isolated for the velocity, which gave me v=2.209 m/s. This is the launch velocity once the box leaves the spring.
Since it is a frictionless surface, this speed would be maintained (and becomes v0) when the box encounters the rough patch.
The work done by friction on the rough patch should thus be equal to the initial kinetic energy of the box. i.e. with speed 2.209 m/s, KEi = 7.15 J, and W by friction= Friction * d.
Using the coefficient of kinetic friction and the normal force, kinetic friction is 3.47 N, which gave me a d (how far the box travelled before coming to a stop) to be 2.06 m, which was wrong.
Where am I going wrong?

2. Mar 1, 2012

### Staff: Mentor

Something's fishy with your numbers. Can you show your calculation for the KE of the launched box?

3. Mar 2, 2012

### shrutij

KE=Spring Potential energy
1/2mv^2=1/2 kx^2. I used this to find vf, which i then used to find KE.
so KE=1/2*2.93*2.209^2= 7.15 J.
This should be the initial KE of the box when it encounters the rough patch. Therefore, due to conservation, KEi=KEf + W done by friction. Since the final KE is 0 (box comes to rest), KEi=W
Since W= friction * d, 7.15=friction*d. I found the force of friction using the coefficient*m*g= 3.47. Therefore d=7.15/3.47=2.06m, which is wrong.

4. Mar 2, 2012

### Staff: Mentor

I don't understand. If the potential energy (1/2)kx2 is equal to the kinetic energy, why calculate the velocity first and then convert back to kinetic energy? Also, something's gone wrong in your math along the way. The value you ended up with for KE looks way too big.

5. Mar 2, 2012

### shrutij

Am I right in assuming that all the spring potential energy should be converted into KE of the box?

6. Mar 2, 2012

### Staff: Mentor

Yes, of course.

7. Mar 2, 2012

### shrutij

Great, so going from that, 1/2kx^2=KE of the box.
This gives me the same answer I got before. 1/2*92.2*0.155 = 7.15 J.
I'm not sure what I'm missing in my calculations if this KE appears too big.

8. Mar 2, 2012

### LawrenceC

It's x squared not x to the first power.

9. Mar 2, 2012

### shrutij

Well, I feel silly....
i'll just be happy that atleast I understood the physics right. :)
thanks!