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How much force do I exert on the Earth?

  • Thread starter cs44167
  • Start date
4
0
Homework Statement
Calculate your acceleration while jumping, the net force that acts on you while jumping, and the force you exert on Earth when you jump.
Homework Equations
w = mg

acceleration = Net force / mass

2.2 lb = 1 kg
This question required measurements which are the following:

my weight = 108 lbs
crouching distance (the distance from my regular height to where I crouch) = 90.6 cm
jump height = 60.4 cm

I first converted lb to kg, and I got 49.09 kg. I then used this value for w = mg and inputted 9.80 for g and got w = 481.09 N. I then used this force for a = net force / mass and found the acceleration to be 9.7999 m/s/s.

This answer was wrong, and I was wondering why I went wrong in my work. The answer made sense to me — the only force, neglecting air resistance, acting on someone in the middle of their jump is acceleration due to gravity.
 

BvU

Science Advisor
Homework Helper
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Hello Cesium, :welcome: !

Doesn't your result (9.7999 m/s) look a lot like ##g## ?

(Small wonder if you first do ##W = mg## and then ##a = W/m## :smile: )​

I envy your 60 cm !

Jumping has three phases: you go from crouching to right up (accelerating), then you fly (basically free fall) and when you land you crouch again to soften the landing (decelerating).

This answer was wrong,
That surprises me: I agree with your reasoning and with your result. How do you know it's wrong (do you have the right answer, or just a red cross or something ?)

Perhaps - if your teacher is a computer in a strange backward country, the answer is expected to be expressed in antique units ? (Such as lbf, feet/forthnight2 or other :wink:)
 
4
0
Hello Cesium, :welcome: !

Doesn't your result (9.7999 m/s) look a lot like ##g## ?

(Small wonder if you first do ##W = mg## and then ##a = W/m## :smile: )​

I envy your 60 cm !

Jumping has three phases: you go from crouching to right up (accelerating), then you fly (basically free fall) and when you land you crouch again to soften the landing (decelerating).

That surprises me: I agree with your reasoning and with your result. How do you know it's wrong (do you have the right answer, or just a red cross or something ?)

Perhaps - if your teacher is a computer in a strange backward country, the answer is expected to be expressed in antique units ? (Such as lbf, feet/forthnight2 or other :wink:)
It’s on a website and gives a red x when the answer is incorrect. There’s never been an issue and the answer for acceleration is to be expressed in m/s/s. The only thing I was thinking was since acceleration is a vector if not having a negative sign was the issue.

We have three submissions, the first I put 9.80 m/s/s, then -9.80 m/s/s, and now I’m stuck.
 

BvU

Science Advisor
Homework Helper
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I'm afraid I am stuck as well. Doesn't help but perhaps feels a little better.

I don't suppose the website is accessible for pagans from outside ?
 
664
251
I'm afraid I am stuck as well. Doesn't help but perhaps feels a little better.

I don't suppose the website is accessible for pagans from outside ?
The correct answer is a maximum upward acceleration a=.67g. This is what is required to produce a jump height of 60.4cm with an acceleration stretch over a distance of 90.6cm. Then add in the weight.
 

BvU

Science Advisor
Homework Helper
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I didn't read that in the problem statement, but I can understand that it's somewhat implicitly intended (semantics?). So a jump ends in that interpretation once the object leaves the ground.

@cs44167: If your career doesnt depend on it or you have more tries left, give it a try !

@hutchphd : you seem to know the website ?
 
664
251
Perhaps I overstated....I think the correct answer is 1.67W....... I don't know the website. Be glad to help more tomorrow.
 

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