Understanding Work and Energy in a Trampoline Jump

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In summary, the conversation discusses a problem involving a gymnast jumping on a trampoline and reaching a maximum height before falling back down. The question asks for the initial speed with which the gymnast leaves the trampoline, and the conversation explores different equations and concepts to solve for this. It is determined that the work done by gravity is equal to the change in potential energy, and the direction of this work depends on the initial and final heights. It is also clarified that if there are no non-conservative forces acting, the work done is equal to 0.
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Maxo
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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


[tex]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)[/tex]

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:

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

[tex]Wnc = -1/2 \cdot m \cdot V_0^0 + m \cdot g \cdot h_f - m \cdot g \cdot h_0[/tex]
Solving for V0, we get:
[tex]V_0=\sqrt{2 \cdot g (h_f - h_0) - 2Wnc/m}[/tex]
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 [tex]W_gravity = m \cdot g (h_0 - h_f)[/tex]

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 [tex](h_0 - h_f)[/tex] instead of [tex](h_f - h_0)[/tex]?

Please help me clarify this so that I can understand these concepts.
 
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  • #2
Maxo said:
Now, what is Wnc?
What does Wnc mean? Do any non-conservative forces act here?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
Maybe my second question can be clarified though? Why would they write (h0-hf) instead of (hf-h0)?
 
  • #7
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.
 
  • #8
Doc Al said:
The work done by gravity is -ΔPE.

Interesting. Could you please explain why it's not ΔPE?
 
  • #9
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?
 

FAQ: Understanding Work and Energy in a Trampoline Jump

What is the purpose of a gymnast using a trampoline?

The purpose of a gymnast using a trampoline is to train and improve their skills for performing aerial movements and acrobatics. This includes building strength, coordination, and body control.

What are the potential risks associated with using a trampoline as a gymnast?

The potential risks associated with using a trampoline as a gymnast include falling or landing incorrectly, leading to injuries such as sprains, fractures, or concussions. Injuries can also occur if the gymnast attempts complex movements without proper training and supervision.

How does a trampoline benefit a gymnast's overall performance?

A trampoline can benefit a gymnast's overall performance by helping them develop power, balance, and spatial awareness. It can also assist in learning new skills and perfecting techniques by providing a controlled and consistent environment for practicing.

What are some common techniques used by gymnasts on a trampoline?

Some common techniques used by gymnasts on a trampoline include somersaults, flips, twists, and various body positions such as tucks, pikes, and straddles. These techniques require a combination of strength, timing, and control.

Can a trampoline be used for competitive gymnastics?

Yes, trampolining is a recognized sport in which athletes compete in events such as individual and synchronized trampoline, double mini trampoline, and tumbling. It is also a component of artistic gymnastics, where trampoline routines are performed as part of a floor exercise.

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