# Work, Energy

1. Aug 5, 2015

### Dalip Saini

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

You hold a 500N barbell 0.5m above a level floor for 100s . During this time, the amount of work you do on the barbell is

• A

1000J

• B

0J

• C

500J

• D

50000J
2. Relevant equations

W = Fdcos(phi)
3. The attempt at a solution
F= 500 N
d = 0.5 m
Since the difference in the vector direction is 0, I thought that all you had to do 500*0.5 = 250 J because cos(0) equals 1. But the answer is zero and I don't understand why.

2. Aug 5, 2015

### haruspex

Equations aren't much use unless you know what all the variables in it represent.
What is d in this equation? Yes, I know it's a distance, but what distance in relation to F, exactly?

3. Aug 5, 2015

### Dalip Saini

Its the distance of the barbell above the ground. So is that vector pointing down and the force is pointing up? But still cos(180) is still -1 not 0

4. Aug 5, 2015

### haruspex

Forget this specific problem for the moment. You quoted a standard equation. How is d defined in that equation?

5. Aug 5, 2015

### Dalip Saini

the displacement?

6. Aug 6, 2015

### haruspex

Displacement of what?