# Homework Help: Calculate Work

1. Feb 27, 2014

### yazz912

1. The problem statement, all variables and given/known

A constant Force= <2,4,1> (in Newtons) moves an object in space from (0,0,0) to (2,4,5) ( distance is measured in meters) Calculate the Work done.

2. Relevant equations

W= || proj of F onto pq|| ||PQ||

W= F• vector PQ

3. The attempt at a solution
Well I'm guessing for starters I'd have to make a vector component from (0,0,0) to (2,4,5)
So I know PQ would be =< 2,4,5>

I am stuck on what my next step would be. W= F•d when the force is acting in line of motion with the object. But that is not the case.

2. Feb 27, 2014

### Ray Vickson

The way you have written it is correct, whether or not W acts along d (assuming that W and d are both vectors).

3. Feb 27, 2014

### yazz912

So if I have two different vectors being displacement and force.
I dot them to calculate work? Would it really be that simple?

4. Feb 27, 2014

### tiny-tim

5. Feb 27, 2014

### Ray Vickson

Yup. That's what your textbook will tell you.

6. Feb 27, 2014

### yazz912

Wow. I really assumed i needed to do more "work" to solve it. ;)

Therefore W should = 25 Nm

7. Feb 27, 2014

### yazz912

Would I not have to find || PQ|| and ||F|| and then dot?

8. Feb 27, 2014

### Ray Vickson

Why would you do that? How would you compute the dot product if you could do that? Would you not need the angle between the two vectors? How would you get that?

I am amazed that you are even asking the question---you are taking something simple and making it hard.

9. Feb 27, 2014

### yazz912

I guess I probably over think the problem, considering this is part of my take home test I assume it isn't going to be as easy as it seems..

Reason I asked was bc on all my notes and in the book it shows W= ||F|| ||PQ||

10. Feb 28, 2014

### tiny-tim

(just got up :zzz:)

both formulas are correct (and give the same result)

which you use depends on what information you're given

in this case, you're given the coordinates, so it's easier to use b1c1 + b2c2 + b3c3

but if instead you were given the magnitudes and the directions, it would be easier to use ||b|| ||c|| cosθ

11. Feb 28, 2014

### yazz912

Ohhh ok that answers my question thank you so much tiny-tim! more or so for not making me feel like an idiot for asking that last question lol :)