Question about net work and displacement

  • #1
FQVBSina
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I know that by physics definition if displacement is zero, work is zero.

However, if I push an object 5 m to the east, and then move to the other side of the object and push it 5 m back to the west. I think in this case I have always done positive work on the object and hence the total work done shouldn't be zero.

If I swing the object around in a circle then my force is perpendicular, so the work done is zero, but what about if I pushed a cart around in a circle from behind?
 

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  • #2
Ibix
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I think in this case I have always done positive work on the object and hence the total work done shouldn't be zero.
When it stopped moving on each leg, which way was the force that stopped it pointing?
 
  • #3
FQVBSina
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When it stopped moving on each leg, which way was the force that stopped it pointing?
Stopped by friction. As in if I stop pushing then it doesn't move. And friction points in the opposite direction. But friction is a dissipative force
 
  • #4
Ibix
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Stopped by friction. As in if I stop pushing then it doesn't move. And friction points in the opposite direction. But friction is a dissipative force
So the total work done by friction has the opposite sign to the work you did. What's the magnitude of the work it did compared to the work you did?
 
  • #5
FQVBSina
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So the total work done by friction has the opposite sign to the work you did. What's the magnitude of the work it fid compared to the work you did?
Ok, I still don't know why you are asking about friction since friction would just be calculated into the net force. But the friction's work would be whatever the friction force is times the total distance (not displacement since friction is not conservative), which will be less than the pushing force.
 
  • #6
Ibix
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But the friction's work would be whatever the friction force is times the total distance (not displacement since friction is not conservative), which
So you are saying that you do work on the object, implying that it gains kinetic energy. And the frictional force does work of the opposite sign and a smaller magnitude. So the object is left with kinetic energy - it hasn't stopped. How are you going to stop it?
 
  • #7
FQVBSina
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So you are saying that you do work on the object, implying that it gains kinetic energy. And the frictional force does work of the opposite sign and a smaller magnitude. So the object is left with kinetic energy - it hasn't stopped. How are you going to stop it?
I am going to stop pushing on it so the friction dissipates the energy and it stops. Then I move to the other side and push the object back.
 
  • #8
Ibix
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I am going to stop pushing on it so the friction dissipates the energy and it stops. Then I move to the other side and push the object back.
Right. So the frictional force does the same magnitude of work as you do, but with the opposite sign. So the total work on the object is?
 
  • #9
FQVBSina
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Right. So the frictional force does the same magnitude of work as you do, but with the opposite sign. So the total work on the object is?
The total work is (myforce - friction) x 5 + (myforce - friction) x 5
 
  • #10
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I know that by physics definition if displacement is zero, work is zero.
This is not a general definition, it is only the definition for the specific case of a constant force. In your description the force is not constant.

The general formula is ##W=\int_S F \cdot ds ##
 
  • #11
FQVBSina
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This is not a general definition, it is only the definition for the specific case of a constant force. In your description the force is not constant.

The general formula is ##W=\int_S F \cdot ds ##
There we go! This is what I was looking for. Work integral is path dependent. Thanks!
 
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  • #12
Ibix
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The total work is (myforce - friction) x 5 + (myforce - friction) x 5
And that is? (Hint: it starts with zero kinetic energy, it ends with zero kinetic energy - what's the change in energy?)
 
  • #13
Delta2
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Not only for constant force but for a conservative force, the total work is zero if the total displacement is zero. This is a consequence of the definition of the conservative force and the definition of work and the gradient theorem.

In the general case a force might not be conservative, the gradient theorem cannot be applied, so the work becomes path-dependent,so the total work is not necessarily zero even if the total displacement is zero.
 

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