How Is Work Calculated When a Force Varies with Position?

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Homework Help Overview

The problem involves calculating the work done by a force that varies with position along the x-axis, described by the equation Fx = 10x - 2.0x² N. The original poster is tasked with determining the work done as the object moves from x = -1 to x = +6.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the work done by integrating the force equation but expresses confusion over the result not matching the provided options. Some participants point out the need for proper integration and the correct application of limits.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach. There is recognition of the need to correctly apply integration and limits, but no consensus has been reached on the correct method or answer.

Contextual Notes

Participants are addressing potential misunderstandings regarding the integration process and the limits of integration, which may affect the calculation of work done.

b-pipe
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Homework Statement


A force acting on an object moving along the x-axis is given by Fx = 10x-2.0x2 N where x is in m. How much work is done by this force as the object moves from x = -1 to x= +6

Distance = 7 m

Homework Equations


Work = Force * Distance cosine theta

The Attempt at a Solution


The equation for force is already given so I found force by integrating the equation Fx = 10x-2.0x2 to get Fx = 10-4.0x then I substituted the x for 7 to get -18 Joules. The problem I have is that the answer I got isn't among the options.
 
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You differentiated instead of integrated.
 
and Put the limits for x properly from -1 to +6, (Wx=+6 - Wx=-1)
 
Yuqing said:
You differentiated instead of integrated.

mukundpa said:
and Put the limits for x properly from -1 to +6, (Wx=+6 - Wx=-1)

Oh, I see thanks.
 

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