Can You Prove Conservation of Energy in Vertical Stone Throwing?

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Homework Help Overview

The discussion revolves around proving the conservation of energy in the context of a stone thrown vertically upward, considering the effects of air drag. The original poster presents equations related to the maximum height and speed before impact, seeking assistance in connecting conservation of energy with kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between conservation of energy and kinematic equations, with some attempting to derive acceleration using Newton's second law. Questions arise regarding the integration of these concepts and the implications of forces acting on the stone.

Discussion Status

Some participants have provided insights into deriving acceleration and relating forces, while others express confusion about the connections between the equations. There is an ongoing exploration of the problem without a clear consensus on the approach to take.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. The discussion reflects uncertainty about the application of equations and the effects of air drag on the stone's motion.

jenavira
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This is driving me nuts.

A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is

h = v(i)^2 / 2g(1+f/w)

and (b) show that the stone's speed just before impact with the ground is

v = v(i) (w-f/w+f)^1/2


I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.
 
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find the stone's acceleration using Newton's 2nd law.
 
Huh. I'm getting closer. But I still have no idea how I'm supposed to relate the conservation of energy and kinetic motion equations. (It's probably something really obvious, and I'll feel like an idiot for not realizing it sooner, but I've been staring at these for so long...)
 
what is the acceleration that you have got ?
 
Well, F = ma so a = F/m...and the force in question would be the initial velocity. But I'm not at all sure how that helps...
 
If upward direction is taken to be positve than using Newton's 2nd law you get
-f+(-W)=ma but since W=mg than m=W/g
then you get a = -g(1+f/w) and using this constant acceleration substistute into
[tex]V^2=u^2+2as[/tex] to get the max height. becareful with the sign.
 
...right. I'm going to sound like an idiot, but could you go through that accelleration in a bit more detail? I'm still a bit confused.
 
[tex]\Sigma\vec{F}=m\vec{a}[/tex] ,right /
two downward forces which are the stone weight and the drag force act on the stone when it is moving upward ,right ?
if the upward direction is chosen to be positive, then downward direction must be negative. then we have -W for the stone weight and -f for the drag force. negative signs indicate the direction and W and f are their magnitudes.
you know that the magnitude of the stone weight is given by W= mg W, m and g are all positive because i am talking about magnitudes.
then substitute into the equation above , you will get what i get.
 
I think I've managed it now. Thanks a lot. :)
 

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