How Much Work is Done by Gravity?

  • #1
84
1

Homework Statement


2.png

The above image shows Yelena Isinbayeva (64.1 kg) 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.

How much work was done be gravity during the time the vaulter went from 2.48 m above the ground to her peak height?

Homework 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

The Attempt at a Solution


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I tried:
W=FΔd
W=(1778.33)(5.32-2.48)
W=5050.46

PEg=mgh
PEg=(64.1)(9.81)(5.32-2.48)
PEg=1785.85

Ws=(1/2)(1090)(5.32)(2.48)2
Ws=4395.75

These answers are all incorrect. I'm not sure how to approach this problem in another way.
 
Last edited:

Answers and Replies

  • #2
7
2
Just observing U R attempts at sol'n:
The only one that makes sense to me, according to the question asked is the 2nd attempt using PEg = mgh. The force of gravity being "mg" the weight of Y.I. and h being her vertical displacement. Now then this computed value has UNITS of J and can only be significantly reported to THREE (3) digits because of the significance of the given data. In addition to that the work done by gravity is NEGATIVE because the force of gravity (in this case) acts in a direction 180° from the displacement.
Work = Force • Displacement • Cos Θ = (628.8)(5.32-2.48)(Cos 180°) = -1785.8 ≈ - 1790 J ANS <= see if that isn't correct :>)
 
  • #3
84
1
Just observing U R attempts at sol'n:
The only one that makes sense to me, according to the question asked is the 2nd attempt using PEg = mgh. The force of gravity being "mg" the weight of Y.I. and h being her vertical displacement. Now then this computed value has UNITS of J and can only be significantly reported to THREE (3) digits because of the significance of the given data. In addition to that the work done by gravity is NEGATIVE because the force of gravity (in this case) acts in a direction 180° from the displacement.
Work = Force • Displacement • Cos Θ = (628.8)(5.32-2.48)(Cos 180°) = -1785.8 ≈ - 1790 J ANS <= see if that isn't correct :>)
Thank you! This was correct (my professor "doesn't care" about sig. figs. and just takes our answers to two decimal places so -1785.80J worked. I was confused before because he said there is no such thing as negative energy so I accidentally was using that for the work & we weren't given that W=FΔdCosθ, so thank you for showing me! :)
 
  • #4
84
1
@Jim Kadel (Or anyone else!) Do you think you could help with this one too? Or point me in the right direction?

If the vaulter first makes contact with a landing mat when her center of mass is 1 m above the ground, then how long did it take her to fall from her peak height to the mat's surface?

I tried this:

v=d/t
t=d/v
t=5.32/9.81
t=0.54s

v=d/t
t=d/v
t=4.32/9.81
t=0.44s

This is incorrect and I know that 9.81 is an acceleration not a velocity, but I don't know how to go about getting the velocity without time (or how to go about the question in a different way).
 
  • #5
84
1
Never mind I got it! Needed to use a projectile equation: t=(2d/g)1/2
 
  • #6
84
1
@Jim Kadel Okay, I only have two questions left and they're pretty much the same & I'm not sure what I'm doing wrong here...

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?

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 bring the vaulter to a complete stop. What was the average force exerted by the mat, on the vaulter, over the 0.5 m?

If someone could help me out please I would really appreciate it!!!
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=9.43.23

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

Tried it negative the second time just to see...
 
  • #7
1
0
Quick question, how did you manage to find peak height?
 
  • #8
CWatters
Science Advisor
Homework Helper
Gold Member
10,533
2,298
When the vaulter lands on the mat he has some KE as well as PE.
 

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