Problem: Work done by a Variable Force

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Homework Help Overview

The discussion revolves around calculating the work done by a variable force on an object, given its initial position and specific work values. The subject area includes concepts from mechanics, particularly relating to work and force as functions of position.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between force and position, questioning the use of a single force value in calculations. There are attempts to derive the final position based on given work values, with some participants expressing confusion about the implications of negative work.

Discussion Status

Participants have engaged in a back-and-forth discussion, clarifying the nature of the force as variable and addressing the calculations needed for both positive and negative work scenarios. Some guidance has been offered regarding the interpretation of the force graph and its impact on the calculations.

Contextual Notes

There is mention of specific work values (0.21 J and -0.19 J) and the initial position of the object (0.40 m), with participants noting the need to consider the variable nature of the force as position changes. The discussion reflects uncertainty about how to approach the calculations for negative work.

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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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