Problem: Work done by a Variable Force

  • Thread starter lucky_star
  • Start date
  • #1
33
0

Homework Statement


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.

2pt3yue.jpg


Homework Equations


W = F1x1 + F2(x2 - x1)


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:

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,306
47
How do you determine the work done from the graph you have given? Does that correspond to the formula you gave?
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,836
251
Welcome to PF!

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

but i DON'T get the answer correctly.

Hi lucky_star! Welcome to PF! :smile:

Why have you used only one value of F?

F has different values as x changes.
 
  • #4
33
0
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
33
0


Hi lucky_star! Welcome to PF! :smile:

Why have you used only one value of F?

F has different values as x changes.

I know the initial position is .40m, but I don't know which are the corresponding forces :(
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,836
251
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

Noooo … :cry:

Look at the graph!

The force is 0.8N only for x ≤ 0.50m.

Try again. :smile:
 
  • #7
33
0
Noooo … :cry:

Look at the graph!

The force is 0.8N only for x ≤ 0.50m.

Try again. :smile:

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
tiny-tim
Science Advisor
Homework Helper
25,836
251
:biggrin: Woohoo! :biggrin:

Have a free quark! o:)

(and now try b … :wink:)
 
  • #9
33
0
do I just simply plug in the value of w= -0.19J and then solve it, or else?
 
  • #10
tiny-tim
Science Advisor
Homework Helper
25,836
251
do I just simply plug in the value of w= -0.19J and then solve it, or else?

You tell me! :wink:
 
  • #11
33
0
You tell me! :wink:

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
tiny-tim
Science Advisor
Homework Helper
25,836
251
I think it not a substitution, because work have a negative sign right now. So, i am not sure what to do :(

Hint: work = force x distance,

so if work is negative, then … ? :smile:
 
  • #13
33
0
Ok. I got to run to class now, I'll try in class. Thank you so much for helping me. Have a nice day.
 

Related Threads on Problem: Work done by a Variable Force

  • Last Post
Replies
6
Views
15K
  • Last Post
Replies
3
Views
18K
  • Last Post
Replies
4
Views
3K
Replies
8
Views
1K
  • Last Post
Replies
4
Views
901
Replies
4
Views
2K
  • Last Post
Replies
1
Views
4K
Replies
4
Views
10K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
4K
Top