# How much work is done?

1. Dec 7, 2015

### milla7

1. The problem statement, all variables and given/known data
Question
A 7.6 kg body is at rest on a frictionless horizontal air track when a constant horizontal force F with arrow acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force F with arrow is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F with arrow between t = 0 and t = 0.5 s?

2. Relevant equations
x=vot +1/2at^2
v=at
m/2(Vf^2-Vi^2)
3. The attempt at a solution
When I work the problem I get 0.152 Joules.
.8 = 1/2 a(2.0)^2
a=0.4
v=0.4(0.5)
v=0.2
1/2(7.6kg)(.2^2)= 0.152J.

I work it this way and I have no idea why it is wrong. The answer is supposed to be 0.0973.

Last edited: Dec 7, 2015
2. Dec 7, 2015

### Staff: Mentor

Where does .8 come from?
The problem statement is inconsistent - a constant force cannot produce this pattern. There are multiple ways to approach this problem and they will lead to different answers.

3. Dec 7, 2015

### milla7

plug it into 1/2 at^2 = x along with t = 2.0 to get the acceleration. Then I use that to find the velocity at .5 seconds.

4. Dec 7, 2015

### Staff: Mentor

I know, but where does the number 0.8 come from? It is not the position of the object after 0.5 seconds.
Edit: wait, you used the 2 seconds. I guess that's the origin of the discrepancy.

Edit2: Do the same calculation with 0.5 seconds and you get half the other value. Weird.

5. Dec 7, 2015

### milla7

Ah... yea thanks... though something is off with that problem that doesn't make sense... got it to work now though.