# In a Force vs Distance graph, is work the area underneath?

1. Nov 1, 2007

### carmenn

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

In a Force vs Distance graph, is work the area underneath?
A person pushes a shovel into the ground to do some spring gardening. He applies a force to the shovel over the following displacement.

0 N - 0 m
4 N - .02 m
8 N - .04 m
12 N - .06 m

Draw a graph, and find work done by man on shovel over the .06 m.

2. Relevant equations
F x d = W

3. The attempt at a solution
So at first, I calculated all the work, using F x d, and added all of them together, but he said that was wrong. And after we multiplied each of the the force by .02 since its in intervals and then added, but apparently we were wrong again. We have drawn a graph, that is a straight line. Is the area underneath the work? If so, why? Thanks

Last edited: Nov 1, 2007
2. Nov 1, 2007

### hotcommodity

Yes, the area underneath the graphed line is the work, tho' you should be using the x-axis for the distance, and the y-axis for the force(that's simply the common convention). Can you show me how you arrived at yoru answer?

3. Nov 1, 2007

### carmenn

why is distance on the x axis?

well, if i use the area under the graph, then it'd be

distance x force / 2

.06 x 12 / 2

.36 J

But can someone explain why? Since i had come up with 1.12 J when adding them all together, and .48 J when using the interval method

4. Nov 1, 2007

### hotcommodity

I'm not particularly sure how you came up with 0.36J and 1.12. Can you show me how you arrived at those answers both mathematically and with words? Keep in mind that the equation $$W = Fcos(\theta)\Delta x$$ is used to find the work done by a constant force. From 0 to 0.06m, the force varies, and this is why we use the interval method. Does that make sense?

5. Nov 1, 2007

### carmenn

I'm actually not sure what I did, but after reading what you wrote and doing a bit of asking around, I finally have the answer. Thank you for your help! Much appreciated!