Work Equation: Force, Displacement, and theta Explained

Click For Summary

Homework Help Overview

The discussion revolves around the work equation W=Fdcos(theta) and its application in different scenarios involving forces acting on a block. Participants explore the implications of applied force versus gravitational force in calculating work done, particularly in the context of friction and displacement.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the work equation in relation to applied forces and gravitational forces. Questions arise about the conditions under which work is calculated as zero or non-zero, particularly in horizontal motion and the angle of displacement relative to the forces involved.

Discussion Status

Participants have shared differing viewpoints on the calculation of work done by various forces. Some have clarified that the context of the question significantly influences the interpretation of work, while others have acknowledged the nuances involved in summing work contributions from different forces.

Contextual Notes

There is an ongoing examination of assumptions regarding the direction of forces and displacement, particularly in scenarios involving horizontal surfaces and the effects of gravity. Participants are considering how these factors affect the overall calculation of work done.

aleksxxx
Messages
23
Reaction score
0
Okay -

This isn't really a homework problem, but i had a question about the work equation, W=Fdcos(theta)

Say, for example, you have a block sitting on a surface and there is friction, but we are not talking it into account.

There is a force of 10N applied parallel to the displacement of the box, which is 10m

I thought the work done would be 100N, because the Cos between displacement and force is 1.

My Kaplan instructor was saying that in that situation the work done would be zero, because you are using F=mg and the discplacement of the box and the cos between those are 90 deg. ==> W=mg(d)cos(90)=0

I know its really elementary, but i just could have sworn that F in the equation was force APPLIED.

thanks.
 
Physics news on Phys.org
Both the answers are plausible, but it is difficult to say which is right unless you tell us what the question is. If you were asked to calculate the work done by the applied force, then you are right. If the force was gravity, and the surface was "horizontal" (perpendicular to the direction of gravitational force), then the instructor is right.
 
neutrino said:
Both the answers are plausible, but it is difficult to say which is right unless you tell us what the question is. If you were asked to calculate the work done by the applied force, then you are right. If the force was gravity, and the surface was "horizontal" (perpendicular to the direction of gravitational force), then the instructor is right.

Ok cool - that makes sense.

But to me it seems for most questions - i.e: "how much work is done by the applied force moving a box of mass M a distance of D" you would get a non-zero answer, but if the question actually specified "How much work is done by gravity..." then you would get zero.

Wouldnt it hold true though that if asked just how much work was done, that this would be the summation of the work (gravity and applied force) giving you the same answer that you would get from the bolded example above?
 
aleksxxx said:
Ok cool - that makes sense.

But to me it seems for most questions - i.e: "how much work is done by the applied force moving a box of mass M a distance of D" you would get a non-zero answer, but if the question actually specified "How much work is done by gravity..." then you would get zero.

Wouldnt it hold true though that if asked just how much work was done, that this would be the summation of the work (gravity and applied force) giving you the same answer that you would get from the bolded example above?

Again, assuming horizontal surface, motion parallel to the surface, blah, blah..., then, yes.

But if it were something like a falling object is being pushed horizontally by a force F, then the total work done would include a non-zero contribution by the gravitational force.
 
neutrino said:
But if it were something like a falling object is being pushed horizontally by a force F, then the total work done would include a non-zero contribution by the gravitational force.

Awesome, basically what i thought.
This would be the case because the displacement is off an angle from the direction of mg, correct?
 
aleksxxx said:
Awesome, basically what i thought.
This would be the case because the displacement is off an angle from the direction of mg, correct?

Yes. Off an angle != 90 degrees.
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 41 ·
2
Replies
41
Views
4K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
3
Views
3K
Replies
25
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K