What is the work that he does on his body?

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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]
 

Answers and Replies

  • #2
cnh1995
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Work done by the man(against gravity) would be equal to the change(increase) in his potential energy.
 
  • #3
haruspex
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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.
 
  • #4
SammyS
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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.
 
  • #5
haruspex
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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.
 
  • #6
SammyS
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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.
 
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  • #7
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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.
 
  • #8
haruspex
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W = F cos theta d
In that equation, what is the necessary geometric relationship between F, d and theta in the physical arrangement?
 
  • #9
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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
haruspex
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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.)
 

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