Finding height dropped from Force vs. Time graph

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Homework Help Overview

The problem involves determining the height from which a pumpkin was dropped, based on data collected from a force vs. time graph. The context includes concepts from mechanics, specifically relating to energy conservation and impulse-momentum theory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the relationship between potential and kinetic energy to find height. There is an exploration of the area under the force vs. time graph, with some confusion about its interpretation as impulse rather than total force. Questions arise regarding the application of conservation of momentum and the relevance of impulse in this context.

Discussion Status

Some participants have provided guidance on the concept of impulse and its relation to momentum. There are varying interpretations of how to approach the problem, with attempts to calculate height based on estimated impulse values. The discussion reflects an ongoing exploration of the concepts involved without a clear consensus on the correct approach.

Contextual Notes

Participants note potential misunderstandings regarding the definitions of impulse and momentum. There is also mention of the dramatic context of the problem, suggesting a desire for a more engaging scenario.

drierplease
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Homework Statement



A pumpkin was dropped from a balcony of a skyscraper. The pumpkin happened to land on a sidewalk force sensor and the below data was collected. The pumpkin weighs 2.7kg.

a) From what height was the pumpkin dropped?

b) From what floor of the skyscraper was the pumpkin probably dropped?

upload_2016-11-29_13-40-39.png


Homework Equations



Potential Before = Kinetic After: mgh = (1/2)mv^2

Solving for v gives: v = sqrt(2gh)

a = F/m

Vf = a*t+Vi
Sf = .5*at^2+Vi*t+ Si

The Attempt at a Solution



First I used mgh = .5mv^2 and solved for v.

This gave me v = sqrt(2gh)

I figure that finding the area under the curve of the graph gives the total force on the pumpkin as it is hitting the ground. I haven't done this yet, but I am not too worried about getting an exact answer for this part. I more want to make sure I know how to solve the rest of the problem.

I tried using kinematic equations to solve for height, but I ended up with h being 0, so I think I am not using the right concept and/or equation here.

We are currently studying conservation of momentum, but I don't see how that is helpful here for determining the height the object was initially dropped from.

Hopefully I am missing something obvious?
 
Last edited:
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Hello. Welcome to PF!

The area under the graph does not give you the "total force". It gives you something called "impulse". Review your notes or textbook for information about impulse and how it is related to momentum.
 
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TSny said:
Hello. Welcome to PF!

The area under the graph does not give you the "total force". It gives you something called "impulse". Review your notes or textbook for information about impulse and how it is related to momentum.

Right, I was mistaken. I think I've got it now.

The impulse is equal to the change in momentum over that time period.

Change in momentum = area under the curve = impulse = J

So, using that I solved change in momentum:

2.7(v) - 2.7(0)
2.7v = J
v = J/2.7

I estimated J by using Reimann Sum, got 49.72.

Then I just used conservation of energy, mgh = .5*m*v^2 and got 17.29 meters for h.

Hopefully this is right, I think I didn't quite understand impulse.
 
drierplease said:
I estimated J by using Reimann Sum, got 49.72.

Then I just used conservation of energy, mgh = .5*m*v^2 and got 17.29 meters for h.
I got J = 49.65 Nm and h = 17.24 m.
I'm not sure what floor that was from, but when thinking "skyscraper", it was not very high.
Once again, a problem with more drama and suspense needed. :)
 
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TomHart said:
I got J = 49.65 Nm and h = 17.24 m.
I'm not sure what floor that was from, but when thinking "skyscraper", it was not very high.
Once again, a problem with more drama and suspense needed. :)

Thanks, I'm glad I was close!
 

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