# Simple Work Done by Force Problem

1. May 4, 2010

### tangibleLime

1. The problem statement, all variables and given/known data
How much work is done by the force $$\vec F = (-4.0\hat\imath-6.0\hat\jmath)\;{\rm N}$$ on a particle that moves through displacement $$\Delta\vec{r}=(-3.0\hat\imath+2.0\hat\jmath)\;{\rm m}$$?

2. Relevant equations
$$A_{x}B_{x} + A_{y}B_{y}$$

3. The attempt at a solution
I decided to use the component form of the dot product method for this problem. (Or is the component form an alternative to the dot product?) Anyways, I set F as A and R as B, resulting in the following:

$$A_{x}B_{x} + A_{y}B_{y}$$
$$F_{x}r_{x} + F_{y}r_{y}$$

$$\vec F = <-4, -6>$$
$$\vec r = <-3, 2>$$

$$-4(-3) + (-6)(2)$$
$$\vec C = <12,-12>$$

Then I took the square root of the sum of the sqaures:

$$\sqrt{12^{2}+(-12)^{2}}$$
$$\sqrt{288} \approx 17$$

My answer of 17 was marked incorrect. Where did I go wrong? Any input is greatly appreciated.

I just realized that I must have made some major mistake. I just kind of "created" vector C out of nowhere. Any direction on this problem would be great.

EDIT: Nevermind, I figured out my error. Instead of creating vector C, I should have just stopped there, added 12 and -12, and got the correct answer of zero. Huzzah!

2. May 4, 2010

### ehild

The dot product is not a vector! It is scalar.

ehild