# Homework Help: Work Done by a Landing Mat

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1. Feb 26, 2016

### x2017

1. The problem statement, all variables and given/known data
Pole vaulter Yelena Isinbayeva (64.1 kg) is at a point in her vault where the pole is maximally deflected (Δx = 1.63 m). The pole has a bending stiffness of 1091 N/m and behaves according to Hooke's Law. At this point in the vault, her vertical velocity is 3.25 m/s and she is 2.48 m above the ground.

Assume the vaulter lands on a mat whose top surface is 1 m above the ground and that the mat deforms by 0.5 m in bringing the vaulter to a complete stop. How much work was done on the vaulter by the mat?

2. Relevant equations
P=W/t
work done = change in energy
Hooke's Law: Fs=kΔx
Ws=(1/2)kΔx2
KE=(1/2)mv2
PEg=mgh
SE=(1/2)kΔx2
Total energy = KE+PE+SE
W=FΔd

I have already calculated the following from previous questions
Work done on the pole = 1449.34J
SE stored in the pole = 1449.34J
Force applied to the pole = 1778.33N
KE in vertical direction = 338.53J
PEg = 1559.48J
Peak height of her centre of mass = 5.32m
Work done by gravity from 2.48m above ground to peak height = -1785.80J
Time it took to fall from peak height to mat = 0.94s
The vertical velocity she first contacted the mat at = -9.21m/s

3. The attempt at a solution
My attempts are all giving me the same number so I'm pretty sure I'm thinking about it or interpreting it in an incorrect way.

I tried
W=FΔd
W=[(64.1)(9.81)](1+0.5)
W=943.23

PEg=mgh
PEg=(64.)(9.81)(1.5)
PEg=-943.23

Tried it negative the second time just to see...

Last edited: Feb 26, 2016
2. Feb 26, 2016

### haruspex

You do not need to find the height of the vault, or the speed on hitting the mat. Just consider the total energy at the instant described and the total energy at the end. Where has the difference gone?
I do notice what looks like a distance of 1.5m in your calculation. Where is that coming from?

3. Feb 26, 2016

### x2017

The 1.5m distance is the mat's distance from the ground + how far the mat sinks down (0.5m)

4. Feb 26, 2016

### x2017

Thinking about what you said about total energy I thought maybe this might work:
W=KEf-KEi
W=(1/2)mv2-(1/2)mv2
W=(1/2)(64.1)(0)2-(1/2)(64.1)(-9.21)2
W=0-2718.61
W=-2718.61J

This turned out to also be incorrect.

5. Feb 26, 2016

### haruspex

The surface of the mat starts 1m off the ground. When depressed .5m by the athlete, how far is it off the ground?

6. Feb 26, 2016

### x2017

Ohhhhhhh my goodness 0.5m! Can't believe I overlooked that, I even have it drawn out on my paper...

W=FΔd
W=[(64.1)(9.81)](0.5)
W=314.41J

This is not correct when I put it into my assignment either

I don't understand. I thought that the work required to stop a falling object was the KE, which I tried above and was not correct.

7. Feb 26, 2016

### haruspex

Yes, you have to take all the energy into account. All that was wrong with your previous attempt (I think) is you had the wrong final height from the ground.
Go back to your first post where you calculated all the energies at the given initial instant. Figure out the final total energy (i.e. at a height of 0.5m), and calculate the lost energy.

8. Feb 26, 2016

### x2017

Okay,

TE=PEg+SE+KE
TE=1559.48+1449.34+338.53
TE=3008.82

But make it -3008.82 because it is opposite the direction Yelena is travelling.
My online assignment told me this is correct. Thank you for your help! :)

9. Feb 26, 2016

### haruspex

Ok!

10. Feb 29, 2016

### Jthoms

How did you get 3008.82 for total energy when I did it I got 3347.35

11. Feb 29, 2016

### haruspex

I assume x2017's last post left out part of the working. As noted in the thread, the remaining PE (0.5m above ground) needs to be subtracted from the energy absorbed. That doesn't quite seem to fix up the discrepancy, but it's close. If you are still bothered, please post your working.

12. Feb 29, 2016

### Jthoms

The above image shows Yelena Isinbayeva (60.1 kg) at a point in her vault where the pole is maximally deflected (Δx = 1.60 m). The pole has a bending stiffness of 1000 N/m and behaves according to Hooke's Law. At this point in the vault, her vertical velocity is 4.18 m/s and she is 2.41 m above the ground.

Assume the vaulter lands on a mat whose top surface is 1 m above the ground and that the mat deforms by 0.5 m in bringing the vaulter to a complete stop. How much work was done on the vaulter by the mat?
I had the same question but different numbers.
For PE I got 1420.9
KE=525.04
SE= 1280
so I did the same thing for TE=1420.9+1280+525.04=3226
It wasn't the right answer so I don't know what I'm doing wrong.

13. Feb 29, 2016

### haruspex

At what height does she come to rest?

14. Mar 1, 2016

### Jthoms

Her peak height is 5.47 m and she lands on the mat and is depressed by 0.5m

15. Mar 1, 2016

### haruspex

In this way of solving the problem (comparing total energy at the described instant with total energy at the end) the peak height is irrelevant.
I asked what height she is off the ground at the end.

16. Mar 1, 2016

### Jthoms

the mat is 1m above the ground and she depresses it to 0.5m so I guess she would be 0.5 m off the ground, its the same question as the above one but with different numbers.

17. Mar 1, 2016

### haruspex

yes. Did you take that into account in your energy calculation?

18. Mar 1, 2016

### Jthoms

I'm not sure how you would take in into account, how would you go about doing that?

19. Mar 1, 2016

### haruspex

How did you calculate the PE in post #12?

20. Mar 1, 2016

### Jthoms

Never mind I got it, thanks for the help.

21. Mar 1, 2016

### x2017

Yeah, I didn't add the KE in, just SE & PE sorry about that!
The way I did it wouldn't work for my roommate, which makes me think I just got lucky... Did adding up energies work for you?

22. Mar 1, 2016

### Jthoms

I did peak height minus 0.5 so 5.47-0.5=4.97 then I did the PE=mgh using that new height and I got it