About work done through exercise (push ups)

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SUMMARY

The discussion centers on the physics of work done during push-ups, specifically analyzing the energy expenditure of a 50 kg woman performing this exercise. The calculation shows that each push-up requires approximately 98 Joules of work, leading to an estimate of about 3 push-ups to expend 300 Joules of energy. Participants clarify that while the work done during the upward motion is positive, the downward motion involves negative work, resulting in a net work of zero. This highlights the complexity of biomechanics and energy expenditure in physical activities.

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  • Familiarity with the equation W = F * d (Work = Force x Displacement).
  • Knowledge of biomechanics related to human movement.
  • Basic grasp of scalar quantities and their properties.
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Iwanttolearnphysics
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Homework Statement
A 50kg woman does push-ups by applying a force to elevate her center of mass by 0.2m. Estimate the number of push ups that the woman must do in order to expend 300J of energy.
Relevant Equations
W = Fs
Hello, everyone. I'm not sure if I did this problem right because there's no answer key. I got this question from a high school introductory Physics book. But it's jaded and I accidentally poured coffee on it, so I can't take a picture of the question. I copied it word for word though.

This is what I did, please let me know if I did anything wrong.
  • W = mas
  • W = 50(-9.8)(0.2)
  • W = 98 J
  • 300 J/98 J =3.06
  • So I think it will take approximately 3 push ups for her to use up 500J of energy.
  • I'm unsure if this is correct. What happens as the distance between her and the ground decreases? Does she do positive work while moving towards the ground? Does she do negative work while being lifted by her arms?
  • According to wikipedia, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement.
  • Energy is a scalar quantity, but there is such a thing as negative work, right?
  • If the work done while she's pushing up on the ground is negative, and the work done as she's coming back to the ground is positive, won't that mean that ultimately she didn't get any work done because they'll cancel out?
 
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Iwanttolearnphysics said:
Homework Statement:: A 50kg woman does push-ups by applying a force to elevate her center of mass by 0.2m. Estimate the number of push ups that the woman must do in order to expend 300J of energy.
Relevant Equations:: W = Fs
This is a poor question, because biomechanics is generally more complicated than elementary physics, owing to the way the muscles work. The woman would get tired simply remaining in the "plank" position at the top of a press-up, without moving at all.

The question would do better to ask you to analyse the work done on an object as you raise and lower it. The human body is generally very inefficient, so the external work done is often much less than the energy expended (I'm not even sure that's a well-defined term, but let's assume it's how much energy is produced by her muscles).

Iwanttolearnphysics said:
This is what I did, please let me know if I did anything wrong.
  • W = mas
  • W = 50(-9.8)(0.2)
  • W = 98 J
  • 300 J/98 J =3.06
That's the elementary calculation all right. To do a press-up requires the woman to do about ##98J## work on her own body mass.

Iwanttolearnphysics said:
  • What happens as the distance between her and the ground decreases? Does she do positive work while moving towards the ground? Does she do negative work while being lifted by her arms?
  • According to wikipedia, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement.

On the way down, or when you lower an object, you must do negative work on the object. In other words, your muscles must absorb the energy in some way. This energy doesn't get stored like in a spring, but rather gets dissipated as heat in your muscles. Again, biomechanics is quite complicated.

Iwanttolearnphysics said:
  • Energy is a scalar quantity, but there is such a thing as negative work, right?
  • If the work done while she's pushing up on the ground is negative, and the work done as she's coming back to the ground is positive, won't that mean that ultimately she didn't get any work done because they'll cancel out?
Energy is a scalar quantity, but work is a scalar product (of two vectors), which can be negative or positive.

The work done overall by a press-up and down is, indeed, zero. In an idealised scenario, the energy would start stored as chemical energy in your muscles and end as heat in your muscles. That's approximately the effect of doing press-ups!

This is why your muscles tire, even though you are doing zero work overall: you are expended the chemical energy in your muscles which must eventually be replaced.
 
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If her center of mass is free falling those 0.2 m, she doesn't have to do any muscular effort, until she has to stop the downwards movement.
Normal push ups are done slower than that, and there is energy used to slowdown that downwards movement.
 
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PeroK said:
This is a poor question, because biomechanics is generally more complicated than elementary physics, owing to the way the muscles work. The woman would get tired simply remaining in the "plank" position at the top of a press-up, without moving at all.

The question would do better to ask you to analyse the work done on an object as you raise and lower it. The human body is generally very inefficient, so the external work done is often much less than the energy expended (I'm not even sure that's a well-defined term, but let's assume it's how much energy is produced by her muscles).That's the elementary calculation all right. To do a press-up requires the woman to do about ##98J## work on her own body mass.
On the way down, or when you lower an object, you must do negative work on the object. In other words, your muscles must absorb the energy in some way. This energy doesn't get stored like in a spring, but rather gets dissipated as heat in your muscles. Again, biomechanics is quite complicated.Energy is a scalar quantity, but work is a scalar product (of two vectors), which can be negative or positive.

The work done overall by a press-up and down is, indeed, zero. In an idealised scenario, the energy would start stored as chemical energy in your muscles and end as heat in your muscles. That's approximately the effect of doing press-ups!

This is why your muscles tire, even though you are doing zero work overall: you are expended the chemical energy in your muscles which must eventually be replaced.
Thank you. Yeah, I figured that it would be more complicated than that, and I had to stop using that book anyway because I feel like it's too dated. Appreciate the detailed response. I'd give you a hundred thumbs up if I could!
 
Lnewqban said:
If her center of mass is free falling those 0.2 m, she doesn't have to do any muscular effort, until she has to stop the downwards movement.
Normal push ups are done slower than that, and there is energy used to slowdown that downwards movement.
Thank you! Appreciate the response.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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