# Spring constant/ block question

1. Apr 2, 2009

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

A 187 g block is launched by compressing a spring of constant k=200N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure.

After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Help would be very much appreciated!!!

2. Relevant equations

PE= 0.5kx^2
f=ma
v^2=Vo^2+2as

3. The attempt at a solution

...? no idea!

2. Apr 2, 2009

### LowlyPion

Welcome to PF.

You know that when the spring releases its energy that the block will have kinetic energy. So how much kinetic energy does it have coming off the spring.

Then how much work will the friction do to slow it down? W = F*d.

3. Apr 2, 2009

okay i got the KE to be;
= 1/2 k x^2
KE = 2.25

then work as
W = 0.27 * 0.85
W= 0.2295 J

so.. KE = 2.25
you can say:
2.25 = 1/2 m v ^2
plug in the mass and find velocity as 5.028m/s when the spring returns to equilibrium?
then how do i find where it comes to rest? =S

4. Apr 2, 2009

### LowlyPion

Not quite.

W = f *d = μ*m*g*d = 2.25 J

d = 2.25/(.27*.187*9.8)

But that's longer than .85.

So what to do now?

I can't see your picture, but a raised curve suggests that the kinetic energy left over will have to go to height against gravity?

Last edited: Apr 2, 2009
5. Apr 3, 2009

Here is a picture of the problem i drew. The end is curved so the block will return, but the question asks where the block will end up :S

6. Apr 3, 2009

okay the picture didnt work but basically it has a block mass that is pushed into a spring and then it slides out onto a 85cm surface with friction coefficient 0.27 and at the other end it has a curved end so it will slide back.

7. Apr 3, 2009

### LowlyPion

So then it is set to return again over the area of friction to the spring again? And again if necessary until the energy is exhausted?

8. Apr 3, 2009

yep so at what distance does it stop 0 being at the beginning of the friction surface and 85 being the end

i worked it out to be about 78cm on the fourth or fifth rebound..
you?

9. Apr 3, 2009

### LowlyPion

Of course it matters which direction it is going when it stops as to whether the remainder ends as x from the start or 85 - x from the start of the frictional area.

I get a trifle under .80 from the start, but I did my numbers crudely, so you're probably right.

10. Apr 4, 2009

how did u work it out to be about 80?

11. Apr 4, 2009

### LowlyPion

Maybe the better question is how you got .78?

12. Apr 4, 2009

i divided 5.028 by 4.25 (4 times the length of the friction applied area)
but i dont know how i got 5.028 haha

13. Apr 4, 2009

sorry, i took 4.25 from 5.028, still dont remember how i got it though =(

14. Apr 5, 2009

### LowlyPion

OK well you had the KE of the object from the spring at release point.

You calculated that to be 2.25 J.

Now you know that it will need to be chewed up by friction for it to stop.

W = f *d = μ*m*g*d = 2.25 J

That means total d comes out to 4.55m for me.

A round trip is 1.7m. That goes into 4.55 twice, but not 3 times.

Subtract 3.4 for those trips. You have a remainder of 1.15 m.

Since it is greater than .85 but less than 1.7 you know it's coming back. So you need to figure its distance going back toward the spring, and subtract that from .85.

Good thing I checked it, because I got a little more than .90 last time I scribbled it out somewhere. As you can see this is different.

15. Apr 5, 2009