What is the work that he does on his body?

  • Thread starter Thread starter alaa amed
  • Start date Start date
  • Tags Tags
    Body Work
Click For Summary

Homework Help Overview

The problem involves a scenario where a 75-kg man pushes a crate up a ramp at a constant speed. The task is to calculate the work done by the man on both the crate and his own body while moving the crate up the ramp, which is inclined at an angle of 17.5°.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of work done against gravity and its relation to potential energy. There are differing views on whether the man does work on the crate or on himself, with some suggesting that the work done on the crate is zero due to no change in kinetic energy.

Discussion Status

The discussion is ongoing, with various interpretations being explored regarding the nature of work done by the man and the implications of gravitational forces. Some participants have provided insights into the relationship between work and potential energy, while others question the assumptions made in the problem.

Contextual Notes

There is a noted complexity in determining the work done on the man himself, and some participants express uncertainty about how to incorporate this into the overall calculation. Additionally, there is a reference to a specific numerical answer that does not align with the potential energy considerations discussed.

alaa amed
Messages
19
Reaction score
0

Homework Statement



A 75-kg man pushes a crate 4.5 m up along a ramp that makes an angle of 17.5° with the horizontal. He exerts a force of 510 N on the crate parallel to the ramp and moves it at a constant speed.

Calculate the work done by man to move the crate, in joules. Be sure to include the work he does on the crate and on his body to get up the ramp.

2. Homework Equations

W = F cos theta d

The Attempt at a Solution


I calculated the work that he does on the box using that above equation but could not figure out what the secind part of the question is getting at[/B]
 
Physics news on Phys.org
Work done by the man(against gravity) would be equal to the change(increase) in his potential energy.
 
cnh1995 said:
Work done by the man(against gravity) would be equal to the change(increase) in his potential energy.
I'm sure that is the intent of the question, but strictly speaking he does no work on the crate or on his own body. Work done on a rigid object equates to the gain in its KE, which is zero here. The work done by the man consists of work done on the objects (zero) plus the work done on the gravitational system consisting of those objects and the Earth.
 
haruspex said:
I'm sure that is the intent of the question, but strictly speaking he does no work on the crate or on his own body. Work done on a rigid object equates to the gain in its KE, which is zero here. The work done by the man consists of work done on the objects (zero) plus the work done on the gravitational system consisting of those objects and the Earth.
True, the net work done on the crate is zero. However, that consists of the man doing (positive) work on the crate and gravity doing (negative) work on the crate.
 
SammyS said:
True, the net work done on the crate is zero. However, that consists of the man doing (positive) work on the crate and gravity doing (negative) work on the crate.
Doesn't seem any more reasonable as a way to view it, but if you wish to go that path you'd have to say the man does some work on the crate and some on the Earth, with gravity doing negative work on both.
 
haruspex said:
Doesn't seem any more reasonable as a way to view it, but if you wish to go that path you'd have to say the man does some work on the crate and some on the Earth, with gravity doing negative work on both.
The work that the man does on the Earth is essentially zero.

Added in Edit:

I do agree that the issue of what work the man does on himself is problematical.
 
Last edited:
cnh1995 said:
Work done by the man(against gravity) would be equal to the change(increase) in his potential energy.
If the man's potential energy is added to the work he does on the crate, it does not yield the right answer which in this case is 3.14x10^3 J.
 
alaa amed said:
W = F cos theta d
In that equation, what is the necessary geometric relationship between F, d and theta in the physical arrangement?
 
haruspex said:
In that equation, what is the necessary geometric relationship between F, d and theta in the physical arrangement?
F is the magnitude of the force acting on the system.
d is the magnitude of the displacement of the system.
and theta is the angle between the force vector and the displacement vector?
 
  • #10
alaa amed said:
F is the magnitude of the force acting on the system.
d is the magnitude of the displacement of the system.
and theta is the angle between the force vector and the displacement vector?
Ok, now please post your working.
(I get nearly 3300J.)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
11K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 14 ·
Replies
14
Views
6K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 23 ·
Replies
23
Views
9K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
7K