Work Equation: Force, Displacement, and theta Explained

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SUMMARY

The discussion centers on the work equation W=Fdcos(theta) and its application in physics problems involving forces and displacement. A participant initially believed that the work done by a 10N force over a 10m displacement would equal 100J, as the cosine of the angle between force and displacement is 1. However, a Kaplan instructor clarified that if the force considered is gravitational (mg) acting perpendicular to the displacement, the work done would be zero, as W=mg(d)cos(90)=0. The conversation highlights the importance of specifying the force in question to determine the correct calculation of work done.

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  • Familiarity with vector components in physics
  • Knowledge of gravitational force and its effects
  • Basic proficiency in trigonometry, particularly cosine functions
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aleksxxx
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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.
 
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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.
 

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