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Hi can someone help me with this question about work done and energy transfer. watch

1. A pole vaulter has a weight of 500N. she vaults to a height of 4 m. when she lands her trainers compress by 1cm. calculate the average force acting on her trainer as she is landing.

i got the answer as 498.75N is that correct?
2. (Original post by Mahira_m)
A pole vaulter has a weight of 500N. she vaults to a height of 4 m. when she lands her trainers compress by 1cm. calculate the average force acting on her trainer as she is landing.

i got the answer as 498.75N is that correct?
Hi Mahira

3. (Original post by AstroST)
Hi Mahira

I basically worked out the work done for her to vault up in the first place by using the formula
Work done=force x distance
So 500N x 4m
I calculated that the work done was 20000 J
And then after that I didn't know what to do so I assumed that she will also need 20000 J of energy to fall and she would obviously fall 4m to get back to the ground and since her trainers compressed by 1cm she'd fall 4.01 m
The I rearranged the work done equation to give me the force
Force= work done / distance
So i did 20000/4.01.
4. (Original post by Mahira_m)
I basically worked out the work done for her to vault up in the first place by using the formula
Work done=force x distance
So 500N x 4m
I calculated that the work done was 20000 J
And then after that I didn't know what to do so I assumed that she will also need 20000 J of energy to fall and she would obviously fall 4m to get back to the ground and since her trainers compressed by 1cm she'd fall 4.01 m
The I rearranged the work done equation to give me the force
Force= work done / distance
So i did 20000/4.01.
Hi

The first part of your answer is good. So just before she lands she will have 2000 J of energy (which will be kinetic energy).

You then need to think about what happens to that 2000 J of energy when she hits the ground.

(Edited to change 20000 J with 2000 J - I assume typo in OPs post)
5. (Original post by Mahira_m)
I basically worked out the work done for her to vault up in the first place by using the formula
Work done=force x distance
So 500N x 4m
I calculated that the work done was 20000 J
And then after that I didn't know what to do so I assumed that she will also need 20000 J of energy to fall and she would obviously fall 4m to get back to the ground and since her trainers compressed by 1cm she'd fall 4.01 m
The I rearranged the work done equation to give me the force
Force= work done / distance
So i did 20000/4.01.
I got an answer completely different to what you got.

What i did was assume that all potential energy is converted to kinetic energy just before the vaulter hits the ground.

So this energy has to be dissipated somehow, so average resistive force acting on the shoe has to dissipate this energy.

As work done=force*distance, then force=work done/force.

Therefore force=2000/0.01=200000N

Although at first this seems extremely large, u have to remember that the vaulter is brought to rest in a very small distance in a very small time so we must have a very large force.

If you’re calculated value of 498.75N is correct then the vaulter would continue accelerating towards earth as there would be a resultant force towards the earth.
6. (Original post by Mahira_m)
I basically worked out the work done for her to vault up in the first place by using the formula
Work done=force x distance
So 500N x 4m
I calculated that the work done was 20000 J
And then after that I didn't know what to do so I assumed that she will also need 20000 J of energy to fall and she would obviously fall 4m to get back to the ground and since her trainers compressed by 1cm she'd fall 4.01 m
The I rearranged the work done equation to give me the force
Force= work done / distance
So i did 20000/4.01.
(Original post by AstroST)
Hi

The first part of your answer is good. So just before she lands she will have 20000 J of energy (which will be kinetic energy).

You then need to think about what happens to that 20000 J of energy when she hits the ground.
Isnt the energy 2000J as 500*4=2000, could be a typo in OPs working.
7. (Original post by Shaanv)
Isnt the energy 2000J as 500*4=2000, could be a typo in OPs working.
It is, yes, I just copied the 20000 J without realising! Thanks for pointing that out. Although the 500 N x 4 m is correct. I agree with your answer of 200000 J too.
8. (Original post by AstroST)
It is, yes, I just copied the 20000 J without realising! Thanks for pointing that out. Although the 500 N x 4 m is correct. I agree with your answer of 200000 J too.
Sweet i figured that u must just have copy and pasted but i didnt want to presume.

🍻
9. Yes sorry the 20000 was a typo it was supposed to be 2000 😊. Thankyou for the explanation I understand my mistake now.

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