Confused about kenetic and potential energy problem

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving kinetic and potential energy, specifically focusing on a mass dropped from a height and its subsequent motion, including the effects of air resistance and energy loss during bounces. Participants are exploring the relationships between energy, forces, and motion in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive equations for the speed of the mass at collision and the height reached after the bounce, incorporating air resistance. They express uncertainty about calculating the total distance traveled after multiple bounces.
  • Some participants suggest analyzing energy loss due to friction and its relationship to the initial potential energy of the mass.
  • There is a question regarding the number of bounces required for the ball to come to rest, indicating a need for further exploration of the problem.

Discussion Status

The discussion is active, with participants sharing their thought processes and equations. Some guidance has been offered regarding energy analysis, but there is still uncertainty about the final calculations and the implications of multiple bounces.

Contextual Notes

Participants are working under the constraints of not providing complete solutions and are focusing on understanding the relationships between energy, forces, and motion. The original poster has made progress on parts of the problem but is seeking further clarification on the final distance traveled after many bounces.

mattysimins
Messages
4
Reaction score
0
In this problem a mass m is dropped from a height h onto a flat surface. when it bounces it has the same speed before the collision. There is a constant force of air resistance f acting on the mass as it goes through this motion. (answer in terms of f, m, h, v, and g. )

the first asks how fast is the mass going as it collides.
Wnc=Ef-Ei (Ef=Uf (zero) + 1/2mv^2) (Ei=Ui (mgh) + 1/2mv^2 (initial v=0))
fhcos180=1/2mv^2-mgh
mgh-fh=1/2mv^2
(2h(mg-f))/m=v^2
got that one.

The second asks how high the mass goes after the collision.
fhcos180=mgh-1/2mv^2 (the final and initial switch because now the mass is moving upwards and the "final" in this equation is when the ball reaches the peak.)
1/2mv^2=mgh+fh
mv^2=h2(mg+f)
h=(mv^2)/(2(mg+f))
v^2=(2h(mg-f))/m so...
h(after bounce)=(h(original)(mg-f))/(mg+f)
got that one too.

Now the final part of the problem asks after many bounces the ball stops. the displacement is -h, but it traveled much farther back and forth. what was that distance?

So far all i have is i know that for the ball to come to rest, the velocity and the height must equal zero. (also, Ef and Ei must equal zero.)

I don't have a clue as to what to do here, so could someone give me a push in the right direction? not the whole solution, please.

edit: advanced my solution to the second part
 
Last edited:
Physics news on Phys.org
Try analysing the energy involved. The ball stops when the energy lost to friction equals the initial potential energy of the ball.

AM
 
wow i must have been tired last night.
Wnc=-Ei (had that)
-fd=-mgh (had that)
d=mgh/f

... is that it? if it is I'm going to kick myself.
 
mattysimins said:
wow i must have been tired last night.
Wnc=-Ei (had that)
-fd=-mgh (had that)
d=mgh/f

... is that it? if it is I'm going to kick myself.
Well, you still have to figure out how many bounces that takes.

AM
 
Nah, i should have typed it neater.
"After many bounces the ball stops (just a statement). The displacement is -h, but over the ball traveled a considerably larger distance. What was that distance?"
 
A 10kg box is pulled 10m up an inclineed plane by a force of 100n parallel to the incline. The coefficient of friction is .1 (a) What is the work done by force? (b)what is the block potential energy? (c)what is the increase in block kinetic energy? (d)What is the work done against friction? (e)What is the final velocity?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
7
Views
2K
Replies
11
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
4K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K