# Homework Help: Spring and energy problem

1. Jun 30, 2015

### J-dizzal

1. The problem statement, all variables and given/known data
The block in the figurelies on a horizontal frictionless surface, and the spring constant is 44 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block's displacement, what are (d) the block's position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/ch0/33.gif

2. Relevant equations
Fs=-kx , k=x2/2

3. The attempt at a solution
i'm trying to first solve for final x postion xf so that i can find work. which will be equivalent to final kinetic energy.

2. Jun 30, 2015

### Nathanael

When the block is at x=2/45 meters then the acceleration of the block will be zero, but the problem asks for where the block stops, which implies the velocity is zero.

3. Jun 30, 2015

### SammyS

Staff Emeritus
You really should explain your reasoning for the various steps you're taking.

I see you solved an equation to find some value for x. What is significant about this value you found for x.

4. Jun 30, 2015

### J-dizzal

K=1/2mv^2 where v=0 . but i was not given mass. so i cannot solve for K here.

5. Jun 30, 2015

### J-dizzal

i erased that equation. i realized that Fs is the spring force not the constant force, so it does not apply to hookes law equation.

6. Jun 30, 2015

### Nathanael

Does the mass make a difference?

7. Jun 30, 2015

### J-dizzal

i guess not. so Kf=0 also

8. Jun 30, 2015

### J-dizzal

i dont see how to relate the kinetic energies to solving the displacement of the object.

9. Jun 30, 2015

### Nathanael

If the kinetic energy is zero, what happened to all the energy?

10. Jun 30, 2015

### J-dizzal

its all potential energy?

11. Jun 30, 2015

### Nathanael

Talk to me in math

12. Jun 30, 2015

### J-dizzal

U=potential engery=work=∫F dx
work=2d for the 2N force
work=∫Fsdx, is Fs=44N/m here? i thought that was the sprink constant k in hookes law?

13. Jun 30, 2015

### Nathanael

Right, this is the work done by each force. For KE to be zero, they must ____ ?

On one side you have units of newtons and on the other side the units are newtons/meter, so thats a dead giveaway that the equation can't be correct.

14. Jun 30, 2015

### J-dizzal

ok, that wouldnt make sense.
so we have two work equations which are simultaneous equations at final x position? it wouldnt make sense to set them equal to eachother though.

15. Jun 30, 2015

### Nathanael

Well in order for the kinetic energy to be zero, they must add to zero (so that way the net work done on the block is zero).

Or you can look at it in different words: (but the idea is the same)
The work done by the external force must equal the change in (KE+PE) of the system. (PE is potential energy from the spring.)

16. Jun 30, 2015

### J-dizzal

i got -1/11 but the correct answer is positive. did i do the math right?

17. Jun 30, 2015

### Nathanael

It should be positive. Can you show your work?

18. Jun 30, 2015

### J-dizzal

2d=-22d2
0=-22d2 - 2d
0= 22d2 +2d
0= 2d(11d+1)
0=11d+1
-1=11d
d=-1/11

19. Jun 30, 2015

### Nathanael

From the equation Wspring+Wexternal.force=0 you get Wspring=-Wexternal.force
However the force of the spring acts in the opposite direction as the displacement, thus it is doing negative work (so your equation is missing an extra negative sign)

Other than this you're good.

20. Jun 30, 2015

### J-dizzal

ok thanks.
i dont understand what part d is asking, because the kinetic energy is zero.

21. Jun 30, 2015

### J-dizzal

or is it asking for the value of dx here? or would it be 0+dx?

22. Jun 30, 2015

### Nathanael

The kinetic energy is zero at the end, but on it's way to the end it wasn't zero. The question is asking you what is the maximum value of the kinetic energy and where does this maximum occur.

I have to go for a bit, so I'll leave you to think about how to solve that one.

23. Jun 30, 2015

### J-dizzal

ok thanks again.

24. Jun 30, 2015

### SammyS

Staff Emeritus
In post # 17, Nathanael point out the following:

In other words: $\displaystyle \ W_\text{external}=(KE+PE) \ .$

Solve for KE & find what x makes it a maximum.

25. Jul 1, 2015

### J-dizzal

solving for KE i got -0.182J.
I dont know why x=0 would not the the x that would make it a maximum because that is when the force of the spring is a minimum. please help thanks.