A 55.0 kg pole vaulter running at 9.0 m/s vaults over the bar

[GalacticSnipes]

Homework Statement

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?

Homework Equations

KE = 0.5*m(vavg)^2

The Attempt at a Solution

KE = 0.5*55(9?)^2
I don't really know how to approach this problem...[/B]

 

[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.

 

[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...

 

[TSny]

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...

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?

 

[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?

 

[TSny]

Have you studied the concept of "potential energy"?

 

[GalacticSnipes]

not in depth

 

[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"?

 

[GalacticSnipes]

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

 

[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.

 

[GalacticSnipes]

yes i have studied that. Wnet = ΔK?

 

[TSny]

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

 

[GalacticSnipes]

Gravity?

 

[TSny]

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

 

[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?

 

[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.

 

[JBA]

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

 

[GalacticSnipes]

no

 

[GalacticSnipes]

Wait, Wnet = ∆k

 

[JBA]

Try looking to see if this relationship is in your textbook

 

[TSny]

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

 

[JBA]

Correct

 

[GalacticSnipes]

W = F*h
∆K=W_net
∆K = F*h
2200 = 539*h
h = 4.08
Would this be the answer?

 

[TSny]

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

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

 

[GalacticSnipes]

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

 

[TSny]

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

Just a quibble about your equation ∆K = F*h.
When a force F acts opposite to the displacement d, it is usual to write the work as W = -Fd where F and d are positive numbers representing the magnitudes of the force and displacement. So, when an object moves vertically upward, the work done by gravity would be written W = -Fh. But I guess you took care of the negative sign by thinking of F itself as a negative number.

Anyway, I think your answer is correct as long as you add the appropriate unit.

 

[GalacticSnipes]

OK, thanks for your help. Looking back at it, it seems pretty easy, lol.
Anyway thanks again.