Understanding Work and Energy in a Trampoline Jump

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Homework Help Overview

The discussion revolves around a problem involving a gymnast jumping from a trampoline, focusing on the concepts of work and energy. The gymnast's initial and maximum heights are provided, and the problem seeks to understand the initial speed at which the gymnast leaves the trampoline while considering the principles of work and energy.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between work done by non-conservative forces and gravitational work, questioning the definitions and equations involved. There is an attempt to clarify the meaning of Wnc and its implications in the absence of non-conservative forces.

Discussion Status

Some participants have offered insights into the relationship between kinetic energy (KE) and potential energy (PE) in the context of the work-energy theorem. Questions remain regarding the specific expressions used in the equations and the interpretation of work done by gravity.

Contextual Notes

Participants are navigating through the definitions of work and energy within the constraints of the problem, particularly focusing on the absence of non-conservative forces and the implications for the equations used.

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Homework Statement


A gymnast springs vertically upward from a trampoline. The gymnast leaves the trampoline at a height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. All heights are measured with respect to the ground. Ignore air resistance, determine the initial speed v0 with which the gymnast leaves the trampoline.

Homework Equations


Wnc = 1/2 \cdot m \cdot V_f^2 + m \cdot g \cdot h_f - (1/2*m \cdot v_0^2 + m \cdot g \cdot h_0)

The Attempt at a Solution


I know this can be solved with pure kinematics and I know how to do that, but this is in a chapter on Work and Energy and I want to understand the concepts there, so that's how I want to solve it.

So here's my try:

Wnc = (KE_f-KE_0) + (PE_f-PE_0)
We know that KE_f will be 0, so it can be removed from the equation.
KE_0 = 1/2 m \cdot V_0^0
PE_0 = m \cdot g \cdot h_0
PE_f = m \cdot g \cdot h_f
So we get:

Wnc = -1/2 \cdot m \cdot V_0^0 + m \cdot g \cdot h_f - m \cdot g \cdot h_0
Solving for V0, we get:
V_0=\sqrt{2 \cdot g (h_f - h_0) - 2Wnc/m}
Correct so far?
Now, what is Wnc? According to my book, since only the gravitational force acts on the gymnast in the air, it is the net force, and we can evaluate the work by using the relation W_gravity = m \cdot g (h_0 - h_f)

I don't understand how they get to this? First of all, how is W_gravity related to Wnc? It seems here they are assuming they are the same?

Secondly, why does the equation they use have the expression (h_0 - h_f) instead of (h_f - h_0)?

Please help me clarify this so that I can understand these concepts.
 
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Maxo said:
Now, what is Wnc?
What does Wnc mean? Do any non-conservative forces act here?
 
Doc Al said:
What does Wnc mean? Do any non-conservative forces act here?

No I guess not, only gravitational force. But if you are saying the equation for Wnc doesn't apply. How are KE and PE related to the work being done?
 
Maxo said:
No I guess not, only gravitational force. But if you are saying the equation for Wnc doesn't apply.
So what does Wnc equal if there are no non-conservative forces acting?

How are KE and PE related to the work being done?
Consider the work-energy theorem.
 
Doc Al said:
So what does Wnc equal if there are no non-conservative forces acting?


Consider the work-energy theorem.

I understand now. Wnc = 0. That's great, I spent half an hour trying to figure out something in a problem when I only missed a minor thing like this.
 
Maybe my second question can be clarified though? Why would they write (h0-hf) instead of (hf-h0)?
 
Maxo said:
Maybe my second question can be clarified though? Why would they write (h0-hf) instead of (hf-h0)?
The work done by gravity is -ΔPE.
 
Doc Al said:
The work done by gravity is -ΔPE.

Interesting. Could you please explain why it's not ΔPE?
 
Maxo said:
Interesting. Could you please explain why it's not ΔPE?
Figure it out. What's the definition of work?
 
  • #10
Another relevant question. If there is a non conservative force involved. Let's say the gymnast gets hit by a ball. Is there any difference on the total amount of work being done depending on where/when the external non conservative force acts? If it appears "at" v0/h0 or at vf/hf or in between?
 

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