# Homework Help: Relating position and force to speed

1. Oct 4, 2011

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

The force graphed in figure is applied to a 1.8-{\rm kg} box initially at rest at x = 0 on a frictionless, horizontal surface.
(Graph is attached)

2. Relevant equations

F=ma (?)

3. The attempt at a solution

Don't know how to relate speed to these two scalars..

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2. Oct 4, 2011

### cepheid

Staff Emeritus
You haven't stated what the problem is. At all.

3. Oct 4, 2011

I need to find the speed when position = 5, 10 and 15cm. Sorry.

4. Oct 4, 2011

### SammyS

Staff Emeritus
What level of math goes with your course?

5. Oct 4, 2011

It's physics 2A, I'm not having trouble with math I'm having trouble with finding the units of m/s out of units like force and position.

6. Oct 4, 2011

### SammyS

Staff Emeritus
Well, if it's calculus based, we might give one answer.

Trig. & algebra another.

Elementary algebra. ???

How is work done related to kinetic energy.

7. Oct 4, 2011

I had to take calculus to get into this course.

8. Oct 4, 2011

### cepheid

Staff Emeritus
A simple application of the work-energy theorem should get you on your way.

9. Oct 4, 2011

### SammyS

Staff Emeritus
What does area under the graph represent?

10. Oct 4, 2011

So F=ma can give me the acceleration of the object, I have no units of time to convert to speed here

11. Oct 4, 2011

The area represents work done?

12. Oct 4, 2011

### SammyS

Staff Emeritus
think about force time distance. Look at cepheid's suggestion.

13. Oct 4, 2011

### cepheid

Staff Emeritus
Yes.

14. Oct 4, 2011

I can solve for work done here but that leaves me with units of (kg*m^2)/s^2, I do not know how to get to speed here. This is probably very obvious, but I'm not seeing it.

15. Oct 4, 2011

### SammyS

Staff Emeritus

What's this theorem?

16. Oct 4, 2011

F=ma?

17. Oct 4, 2011

### SammyS

Staff Emeritus
That's force.

Look at the units again. remember, you're looking for speed.

18. Oct 4, 2011

Err, W=KE=1/2mv^2?

Im having a lot of trouble with physics...

19. Oct 4, 2011

### cepheid

Staff Emeritus
No, that is not the work-energy theorem. That is Newton's 2nd Law. The work-energy theorem relates the work done on an object to its change in energy. If you haven't encountered it before (which seems strange) you can always look it up.

By the way, the work, of course, has dimensions of force*distance, therefore its units are newton-metres = joules.

20. Oct 4, 2011

### cepheid

Staff Emeritus
That's pretty much it, but it should be that the work done on the object is equal to its CHANGE in kinetic energy. Therefore W = ΔKE = (1/2)mvf2 - (1/2)mvi2, where vf and vi are the final and initial velocities respectively.