# Finding spring constant K and compression distance

1. Apr 24, 2014

### mpittma1

1. The problem statement, all variables and given/known data

https://scontent-a-sjc.xx.fbcdn.net/hphotos-frc3/t1.0-9/10294335_1407511836191702_8633020308624670643_n.jpg

2. Relevant equations
PE=.5kx^2
F=-kx

3. The attempt at a solution

I honestly just dont know how to approach a problem where you have to find both the spring constant and the compression distance....

Ive tried employing conservation of momentum, ive tried working backwards with M1...

Im pretty sure I need to find an equation that would solve for either K or x and then use that to find the other value.

Am I on the right track?

2. Apr 24, 2014

### SammyS

Staff Emeritus
Conservation of momentum and conservation of energy will each be usefully to solve relevant aspects of this problem.

Can you see how you may find the the velocity that m1 attains after the spring fully expands and before m1 gets to the friction surface?

3. Apr 24, 2014

### mpittma1

Ya i see that;

100N = .5M1V12

so V1 = 10m/s

right?

4. Apr 24, 2014

### paisiello2

Start with conservation of momentum to get v2 in terms of v1.

You can get v1 from conservation of energy.

5. Apr 24, 2014

### SammyS

Staff Emeritus
Not what I had in mind.

100 N is the maximum force exerted by the spring. (1/2) mv2 is Kinetic Energy. Those are not equal. They're totally different physical quantities.

Calculate the amount of work done by friction in bringing m1 to rest.

6. Apr 24, 2014

### mpittma1

Ya i noticed that after i posted.

i will try that though

7. Apr 24, 2014

### mpittma1

wf = ukmgd = (.75)(2)(9.8)(2) = 29.4 J

8. Apr 24, 2014

### mpittma1

would impulse momentum help here?

9. Apr 24, 2014

### SammyS

Staff Emeritus
Not really.

Are you familiar with the Work - Energy Theorem ? It tells you the relationship between work done and change in Kinetic Energy.

10. Apr 24, 2014

### mpittma1

Ya Wnet=ΔKE

So, set 100N = ΔKE of mass 1 and find the initial velocity produced by the spring?

11. Apr 24, 2014

### mpittma1

disregard that last post its the same mistake as the first time.....

12. Apr 25, 2014

### mpittma1

K im getting close;

.5m1v12=wf

v1 = sqrt(29.4) = 5.422

and to find v2

v2=(m1v1/m2) = 1.36m/s

13. Apr 25, 2014

### mpittma1

ugh finally got it!

thanks for all the help!!

14. Apr 25, 2014

### Staff: Mentor

Where did that 29.4 value materialize from?

15. Apr 25, 2014

### mpittma1

Its the value of the work done by friction

Wf=μk*m*g*d

16. Apr 25, 2014

### Staff: Mentor

Ah. I can now see that you worked that out earlier.