Problem: Work done by a Variable Force

1. Oct 23, 2008

lucky_star

1. The problem statement, all variables and given/known data
An object is acted on by the force shown below. What is the final position of the object if
its initial position is x = 0.40m and the work done on it is equal to
(a) 0.21 J
(b) (b) -0.19 J.

2. Relevant equations
W = F1x1 + F2(x2 - x1)

3. The attempt at a solution

a) W= F(x2-x1) = 0.8N* (x2- 0.40m)

but i DON'T get the answer correctly.

Last edited: Oct 23, 2008
2. Oct 23, 2008

CompuChip

How do you determine the work done from the graph you have given? Does that correspond to the formula you gave?

3. Oct 23, 2008

tiny-tim

Welcome to PF!

Hi lucky_star! Welcome to PF!

Why have you used only one value of F?

F has different values as x changes.

4. Oct 23, 2008

lucky_star

Oh, I made a mistake there. if initial position is 0.40m, the force is 0.8N. I am given W= 0.21J, So I will have:

0.21J= 0.8N*0.40m + 0.8N*(x2 - 0.40m)
0.21J= 0.8x2 - 0.32
x2= 0.66m

That's all I have. However, the correct answer that I have been provided is 0.90m. I really don't know how to solve this problem. Please give a hint!

5. Oct 23, 2008

lucky_star

Re: Welcome to PF!

I know the initial position is .40m, but I don't know which are the corresponding forces :(

6. Oct 23, 2008

tiny-tim

Noooo …

Look at the graph!

The force is 0.8N only for x ≤ 0.50m.

Try again.

7. Oct 23, 2008

lucky_star

AHHH! I Think I understand the problem.

W= 0.8N *(0.5m-0.4m) + 0.4N *(0.75m-0.5m) + 0.2 *(x2-0.75m)
0.21J = 0.080+ 0.1 + 0.2x2
0.18 = 0.2x2
x2 = 0.90m

8. Oct 23, 2008

tiny-tim

Woohoo!

Have a free quark!

(and now try b … )

9. Oct 23, 2008

lucky_star

do I just simply plug in the value of w= -0.19J and then solve it, or else?

10. Oct 23, 2008

tiny-tim

You tell me!

11. Oct 23, 2008

lucky_star

When I substitute -0.19J in the equation I did not get the right answer. thus, I think it's not a substitution, because work have a negative sign right now. So, i am not sure what to do :(

p/S: I have class in about 20 more minutes, and this is my last question.

12. Oct 23, 2008

tiny-tim

Hint: work = force x distance,

so if work is negative, then … ?

13. Oct 23, 2008

lucky_star

Ok. I got to run to class now, I'll try in class. Thank you so much for helping me. Have a nice day.