Problem: Work done by a Variable Force

In summary, when work is negative, then the equation will be backwards. So if work is negative, then the equation will be backwards.
  • #1
lucky_star
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:
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  • #2
How do you determine the work done from the graph you have given? Does that correspond to the formula you gave?
 
  • #3
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.
 
  • #4
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


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 :(
 
  • #6
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:
 
  • #7
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
 
  • #8
:biggrin: Woohoo! :biggrin:

Have a free quark! o:)

(and now try b … :wink:)
 
  • #9
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.
 

FAQ: Problem: Work done by a Variable Force

1. What is work done by a variable force?

Work done by a variable force is the measure of the amount of energy transferred when an object is moved by a force that changes in magnitude and/or direction. It takes into account the varying force applied over a distance.

2. How is work done by a variable force calculated?

The work done by a variable force is calculated by integrating the force function over the distance travelled by the object. This can be represented by the equation W = ∫ F(x) dx, where W is the work done, F(x) is the force function, and dx is the distance travelled.

3. What is the difference between work done by a constant force and a variable force?

The main difference between work done by a constant force and a variable force is that with a constant force, the force remains the same throughout the entire distance, while with a variable force, the force changes at different points along the distance. This results in different amounts of work being done.

4. What are some examples of variable forces in real life?

Some examples of variable forces in real life include pushing a shopping cart, throwing a ball, or rowing a boat. In each of these situations, the force applied changes in magnitude or direction as the object moves.

5. How does the angle between the force and displacement affect the work done by a variable force?

The angle between the force and displacement affects the work done by a variable force by changing the amount of force that is acting in the direction of the displacement. When the angle is 0 degrees, all of the force is acting in the direction of displacement, resulting in the maximum amount of work. As the angle increases, the amount of force acting in the direction of displacement decreases, resulting in less work being done.

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