For what value of x will this object first come to rest?

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Homework Help Overview

The problem involves a 2.0-kg body moving along the x-axis with an initial velocity of 5.0 m/s, subject to a force that varies with position, specifically Fx = (–4.0x) N. The objective is to determine the position at which the object first comes to rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the varying force and question the method of simply multiplying force by distance. There is an exploration of integrating the force to find total work, with some participants attempting to reconcile their calculations with the expected answer.

Discussion Status

Some participants have provided insights into the integration of force over distance and the implications of the force's variability. There is recognition of the need for a different approach to calculating work, and some participants express understanding after further discussion.

Contextual Notes

Participants are navigating the complexities of variable forces and the associated work calculations, indicating a need for careful consideration of the problem setup and assumptions about work done.

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Homework Statement


A 2.0-kg body moving along the x-axis has a velocity vx = 5.0 m/s at x = 0. The only force acting on the object is given by Fx = (–4.0x) N, where x is in m. For what value of x will this object first come (momentarily) to rest?

Homework Equations


0.5 mv^2 = KE
F*d = work

The Attempt at a Solution


Since KE = 0.5 * 2 * 5^2 = 25J
and F*d = -4.0x^2
-4x^2 = 25J
x = 2.5 m
This is how I approached it, but the answer is 3.5m. Can anyone help me out?
 
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Since the force changes with x, what will be the work done by the force when the body moves by a distance dx?
Integrate it to find the total work and the expression you get comes out to be different. why?
you can't find the net work done like this:
gijungkim said:
and F*d = -4.0x^2
everything else is correct. Just the left hand side of the equation.
 
AdityaDev said:
Since the force changes with x, what will be the work done by the force when the body moves by a distance dx?
Integrate it to find the total work and the expression you get comes out to be different. why?
you can't find the net work done like this:

everything else is correct. Just the left hand side of the equation.
Oh I got it now
Fxdx = -4.0x dx
W = -2.0x^2 J = 0.5 * 2 * 5^2 J
x = 3.5m right??
Thank you so much!
 
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The force changes as the position of particle changes. So you can't get the value of work done by simply multiplying F with x.
 
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AdityaDev said:
The force changes as the position of particle changes. So you can't get the value of work done by simply multiplying F with x.
Got it now! Thank you for helping me!
 

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