Problem: Work done by a Variable Force

AI Thread Summary
The discussion revolves around calculating the final position of an object acted upon by a variable force, given specific work values. The initial position is 0.40m, and the user initially misapplies the force in their calculations, leading to incorrect results. After guidance, they correctly identify the need to account for varying force values based on position and successfully calculate the final position as 0.90m for a work value of 0.21 J. For the negative work value of -0.19 J, the user is advised to consider the implications of negative work on the object's movement. The conversation emphasizes the importance of understanding variable forces and their impact on work done.
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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:
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How do you determine the work done from the graph you have given? Does that correspond to the formula you gave?
 
Welcome to PF!

lucky_star said:
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.
 
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!
 


tiny-tim said:
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 :(
 
lucky_star said:
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:
 
tiny-tim said:
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
 
:biggrin: Woohoo! :biggrin:

Have a free quark! o:)

(and now try b … :wink:)
 
do I just simply plug in the value of w= -0.19J and then solve it, or else?
 
  • #10
lucky_star said:
do I just simply plug in the value of w= -0.19J and then solve it, or else?

You tell me! :wink:
 
  • #11
tiny-tim said:
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
lucky_star said:
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
Ok. I got to run to class now, I'll try in class. Thank you so much for helping me. Have a nice day.
 
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