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October 22, 2019, 10:05:42 am

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Prerna94

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Forces Extension q
« on: April 25, 2019, 12:03:11 pm »
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Hello pls help with this question 🙏🏽, it’s from the Cambridge Checkpoints 2019 book

It’s Ch 2 Forces, q15.
Jemima is planning a ‘bungee jump’. She will jump off a high tower, attached to her starting position by a strong elastic cord. Before she hits the ground, the cord will stop her downwards motion.

a) Jemima’s mass is 50kg. The bungee cord is designed to stop her moving downwards after she has fallen a distance of 60m. Unstretched, the bungee cord is 40m long. Calculate the force constant (k) of the bungee cord.

Answer: a= 147N/m

Bri MT

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Re: Forces Extension q
« Reply #1 on: April 25, 2019, 12:19:45 pm »
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Hey, you could break this question into parts to make it more manageable

You could consider first from the top of the high tower to 40 metres below the top of the tower (because this is the unstretched length of the bungee cord).
What speed will she be travelling then?
Then look at the second part - from 40 metres below the top of the tower to 60 metres below the top of the tower (because her speed will be 0 m/s at 60 m below)
What acceleration is required for her to stop in time?

Then try to find the spring constant :)

Remember to define whether up or down is + and be consistent with it. If you're still having difficulty please feel free to let us know :)
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Prerna94

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Re: Forces Extension q
« Reply #2 on: July 05, 2019, 10:38:51 pm »
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Thanks for responding!

I got 28m/s as the velocity at 40m and 19.6m/s^2 as the acceleration at 60m.
I then used ma=kx to find the k constant, however I got 49N/m, which is far off from the answer, 147N/m

Could you please tell me where I went wrong?

Calebark

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Re: Forces Extension q
« Reply #3 on: July 05, 2019, 11:28:41 pm »
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Hey! Thanks for attempting it. A quicker way I've found to use this is using energy conservation, where the energy at the starting position is equal to the energy at the bottom of the bungee cord. In VCE Physics, unless stated otherwise, we assume that all energy is converted to elastic potential energy at  the bottom of her jump. So from this energy at top = energy at bottom. Given she's standing from a height, we can use her gravitational potential to calculate her energy at the top. As such, mgh = ˝k(x2)

Remember that given the bungee cord is will stop after a distance of 60m, so we use that as her height. So first step is to calculate gravitational potential energy

Gravitational potential energy = mgh
mgh = 50kg * 9.8ms-2 * 60m = 29,400 J

Note that the cord only stretches for 20m, as stated in the question, so for the elastic potential energy, x = 20m

mgh = ˝k(x2)
∴ 29,400 J = ˝k(20m2)
∴ k = 147 N/m

For these sorts of questions, it might be quicker to use conservation of energy. This is the method VCAA prefer use to use, as shown in past exam solutions (example: 2006 Physics Exam 1, Question 14) :)

« Last Edit: July 06, 2019, 12:04:02 am by Calebark »
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