Average Force

  • Thread starter Roze
  • Start date
  • #1
14
0

Homework Statement


After falling from rest from a height of 30m, a .50 kg ball rebounds upward, reaching a height of 20m. If the contact between ball and ground lasted 2.0ms, what average force was exerted on the ball?


The Attempt at a Solution



What I did was drew a nice little picture with the 30m and the 20m and the little 2ms interval between them.

So I tried to first figure out what the change in acceleration was while the ball was in contact with the ground.
So I found the velocity when the ball hits the ground using first [tex]\Delta[/tex]y=vot+ayt^[tex]^{2}[/tex] and I found that t=2.5 seconds. So then putting that into the equation v=vo+at I got that the v on impact =-24.5m/s.
So then I tried to find the velocity when the ball leaves the ground. So I know that it travels 20m upwards, at that at the top of the 20m v=0. So I plugged those two things into v[tex]^{2}[/tex]=vo[tex]^{2}[/tex] + 2a[tex]\Delta[/tex]x and I solved for vo. I found that vo was 19.79m/s.
So I think at this point I have that when the ball hits the ground, v=24.5m/s, and when the ball leaves the ground v=19.79m/s and that the ball is in contact with the ground for 2ms.
So I said that a=[tex]\Delta[/tex]v/[tex]\Delta[/tex]t and I got that a=-2355m/s[tex]^{2}[/tex].

So my force in those 2ms =ma =.50kg(-2355m/s2) = -1177.

So since it asks for the average force exerted on the ball I think I'm supposed to get some other forces and then average them together with that one. So I said at the top of the 30m F=0 (before the ball is dropped). The ball hits the ground with an acceleration of -9.8m/s[tex]^{2}[/tex] so the force on impact is (.5kg)(-9.8)=-4.9N. At the top of the 20m rebound a is also -9.8 so the force is the same, -4.9.

So at this point I'm feeling like I'm making this much more complicated that it needs to be, but I add all those forces together:

-4.9+-4.9+0+-1177 and divide by 4 = -296N.

Unfortunately this is the wrong answer, so says the answer key and I can't figure out what to do.

Help?
Hopefully trolling through all of my attempts won't be too daunting.

Thanks!
 

Answers and Replies

  • #2
Hey, a more thorough understanding of impulse forces will make this question ten-times easier for you. We know that the change in momentum, if the impulse approximation is valid, is equal to the impulse force. We should also know that impulse is just the time integral of the force applied to an object during a brief interaction that agrees with the impulse approximation. Taking the integral of this force is equivalent to finding the average force exerted over duration of the impulse and multiplying it by dt. Therefore, by finding the change in momentum, and given that you already know what dt is, you should be able to find Favg quite easily.
 
  • #3
Doc Al
Mentor
45,140
1,443
So I found the velocity when the ball hits the ground using first [tex]\Delta[/tex]y=vot+ayt^[tex]^{2}[/tex] and I found that t=2.5 seconds. So then putting that into the equation v=vo+at I got that the v on impact =-24.5m/s.
OK.
So then I tried to find the velocity when the ball leaves the ground. So I know that it travels 20m upwards, at that at the top of the 20m v=0. So I plugged those two things into v[tex]^{2}[/tex]=vo[tex]^{2}[/tex] + 2a[tex]\Delta[/tex]x and I solved for vo. I found that vo was 19.79m/s.
OK.
So I think at this point I have that when the ball hits the ground, v=24.5m/s, and when the ball leaves the ground v=19.79m/s and that the ball is in contact with the ground for 2ms.
So I said that a=[tex]\Delta[/tex]v/[tex]\Delta[/tex]t and I got that a=-2355m/s[tex]^{2}[/tex].
Careful! What's the change in velocity? Direction counts! The initial velocity is downwards, the final velocity is upwards.
 
  • #4
14
0
I thought about that, but when I plug in the velocities: 19.79-(-24.5)/.002 I get a positive acceleration (22145), which doesn't make sense to me because I think the ball should be slowing down not speeding up during the impact.
Also, with that answer plugged into the average of the rest of the Forces I'm still getting the wrong answer.
 
  • #5
14
0
Also, in response to kplooksafter me, I didn't look at this problem from an impulse perspective because it is a question in a chapter before we covered impulses, so theoretically I should be able to do it without any knowledge of impulse. Which is what I attempted, however unsuccessfully.
 
  • #6
Doc Al
Mentor
45,140
1,443
I thought about that, but when I plug in the velocities: 19.79-(-24.5)/.002 I get a positive acceleration (22145), which doesn't make sense to me because I think the ball should be slowing down not speeding up during the impact.
A positive acceleration doesn't necessarily mean "speeding up", it means that the velocity is changing in the positive direction--which is certainly the case here.
Also, with that answer plugged into the average of the rest of the Forces I'm still getting the wrong answer.
What did you get?
 
  • #7
14
0
I got 7378.4N but the book says the answer is 1400N
 
  • #8
Ahh, sry bout that. I read your workings and you're correct in everything except for the signs in front of the velocities in your average acceleration expression (they must be opposite).
 
  • #9
Doc Al
Mentor
45,140
1,443
I got 7378.4N but the book says the answer is 1400N
Neither answer is correct. How did you get your answer?
 
  • #10
14
0
I figured that when they asked for average acceleration they wanted me to pick a bunch of points, find the force at each, and then average them together. Which may be my principle flaw in this problem...

But I said that as the ball is dropped it has a force of (.5)(-9.8)= -4.9m/s2 and that as the ball rises again it has the same force because the acceleration is the same as is the mass. So I added (-4.9+-4.9+22145) and averaged /3.

When I didn't get the right answer doing that I figured that maybe they were just asking for the average force during the ball's contact with the ground, so I forgot about the -4.9's and just did F=ma with (.5)(22145) but that didn't give me the book's answer either.

I have a feeling this whole problem stems from my interpretation of "average force" in the question.
 
  • #11
Doc Al
Mentor
45,140
1,443
I figured that when they asked for average acceleration they wanted me to pick a bunch of points, find the force at each, and then average them together. Which may be my principle flaw in this problem...
Yes, that's an incorrect approach. They are only talking about the short time during which the ball is in contact with the ground.

But I said that as the ball is dropped it has a force of (.5)(-9.8)= -4.9m/s2 and that as the ball rises again it has the same force because the acceleration is the same as is the mass. So I added (-4.9+-4.9+22145) and averaged /3.
No good.

When I didn't get the right answer doing that I figured that maybe they were just asking for the average force during the ball's contact with the ground, so I forgot about the -4.9's and just did F=ma with (.5)(22145) but that didn't give me the book's answer either.
That might not give the book's answer, but it does give the correct answer.

I have a feeling this whole problem stems from my interpretation of "average force" in the question.
That's part of it.

It's easier than you think. You found the average acceleration, to find the average (net) force just multiply by the mass.
 
  • #12
14
0
I love it when the book is lying to me!

Thank you, I'm glad I didn't do everything totally wrong and that makes sense now.

Thanks again!
 
  • #13
Is the book asking for the average force during the impulse or during the whole duration of the problem (ie when the ball was dropped to when it reaches 20m)? Either way the book's answer is wrong, but you might want to clarify with your prof if you'll have to know how to find the average force acting on the ball over the whole course of events...
 

Related Threads on Average Force

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
1
Views
824
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
996
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
4K
Replies
4
Views
2K
Top