# Homework Help: A 55.0 kg pole vaulter running at 9.0 m/s vaults over the bar

1. Feb 13, 2016

### GalacticSnipes

1. The problem statement, all variables and given/known data
A 55.0 kg pole vaulter running at 9.0 m/s vaults over the bar. Assuming that the vaulter's horizontal component of velocity over the bar is 1.0 m/s and disregarding air resistance, how high was the jump?

2. Relevant equations
KE = 0.5*m(vavg)^2

3. The attempt at a solution
KE = 0.5*55(9?)^2
I don't really know how to approach this problem...

2. Feb 13, 2016

### TSny

Welcome to PF!

Compare the KE while running to the KE while passing over the bar. Are they the same? If not, account for the gain or loss of KE.

3. Feb 16, 2016

### GalacticSnipes

so, something like: 0.5*55(9)^2=0.5*55(1)^2
2227.5=27.5
The difference is 2000, but after finding this, I'm not sure what to do...

4. Feb 16, 2016

### TSny

It would be best not to use an equal sign to compare two quantities that are not equal. But you have the right idea. The difference between 2227.5 and 27.5 is not 2000. So, check that.

You can see that there's a "loss" of kinetic energy in going from the ground to the height over the bar. Is there a gain of some other type of energy that makes up for the loss of kinetic energy?

5. Feb 16, 2016

### GalacticSnipes

oh yeah, didn't check that, the difference is 2200. I think that the KE gets transferred to Thermal Energy, but i haven't really learned about that yet, but is there some kind of equation that relates these two quantities?

6. Feb 16, 2016

### TSny

Have you studied the concept of "potential energy"?

7. Feb 16, 2016

### GalacticSnipes

not in depth

8. Feb 16, 2016

### TSny

Hmm. This problem seems to be designed for testing your understanding of conservation of energy. Are you sure you have not covered the concept of "gravitational potential energy"?

9. Feb 16, 2016

### GalacticSnipes

We might have, but just glancingly. We have learned about Power, Force, Work and KE

10. Feb 16, 2016

### TSny

It's possible to work the problem using concepts of work and kinetic energy. Do you know how kinetic energy is related to the amount of work done on an object? This is sometimes called the "work-energy" theorem.

11. Feb 16, 2016

### GalacticSnipes

yes i have studied that. Wnet = ΔK?

12. Feb 16, 2016

### TSny

OK. You can use that. You've already found ΔK. So, think about Wnet. What force is doing work on the vaulter as he rises above the ground?

13. Feb 16, 2016

### GalacticSnipes

Gravity?

14. Feb 16, 2016

### TSny

Yes. Can you determine the work done by the force of gravity?

15. Feb 16, 2016

### GalacticSnipes

W=Fd
W=

F=ma
F=55(9.8)
F=539

d=∆x=0.5at^2
I don't have the time though, so how can i find distance?

16. Feb 16, 2016

### TSny

You don't need time and you don't need to use any constant acceleration formulas. (The acceleration of the vaulter might not be constant.)

Try to get an expression for the work done by gravity in terms of the unknown vertical distance h, the mass m, and the acceleration due to gravity, g. You will need to take into account that the direction of the force of gravity is downward while the vertical displacement is upward.

Last edited: Feb 16, 2016
17. Feb 16, 2016

### JBA

Are you familiar with the relationship between work and kinetic energy?

18. Feb 16, 2016

### GalacticSnipes

no

19. Feb 16, 2016

### GalacticSnipes

Wait, Wnet = ∆k

20. Feb 16, 2016

### JBA

Try looking to see if this relationship is in your textbook

21. Feb 16, 2016

### TSny

Yes. You know ΔK, so if you could express Wnet in terms of the height, then you can solve for the height.

22. Feb 16, 2016

### JBA

Correct

23. Feb 16, 2016

### GalacticSnipes

W = F*h
∆K=Wnet
∆K = F*h
2200 = 539*h
h = 4.08

24. Feb 16, 2016

### TSny

OK. That's essentially correct except that there are some issues with signs.
ΔK stands for the change in K, therefore ΔK = Kfinal - Kinitial. So, is ΔK positive or negative?

Likewise, you need to consider the sign of the work done by the force of gravity.

25. Feb 16, 2016

### GalacticSnipes

W = F*h
∆K=Wnet
∆K = F*h
-2200 = -539*h
h = 4.08
This correct?