# Work Equation

1. May 17, 2007

### aleksxxx

Okay -

This isnt 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, becuase 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.

2. May 17, 2007

### neutrino

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.

3. May 17, 2007

### aleksxxx

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?

4. May 17, 2007

### neutrino

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.

5. May 17, 2007

### aleksxxx

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

6. May 17, 2007

### neutrino

Yes. Off an angle != 90 degrees.