Determine the work done to move the particle

AI Thread Summary
The discussion focuses on calculating the work done on a particle subjected to a varying net force along the x-axis. The force increases linearly from zero to 284 N between x = 0 and x = 3.0 m, remains constant at 284 N from x = 3.0 m to x = 7.0 m, and then decreases to zero by x = 9.2 m. The area under the force versus position graph, which resembles a trapezoid, represents the work done. Participants emphasize that work can be calculated as the integral of force over distance, confirming that the area under the graph provides the solution. The conversation concludes with a participant expressing satisfaction after resolving their confusion.
Bones
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Homework Statement


The net force exerted on a particle acts in the positive x direction. Its magnitude increases linearly from zero at x = 0, to 284 N at x = 3.0 m. It remains constant at 284 N from x = 3.0 m to x = 7.0 m and then decreases linearly to zero at point A where x = 9.2 m. Determine the work done to move the particle from x = 0 to x = 9.2 m by graphically determining the area under the F versus x graph.


Homework Equations





The Attempt at a Solution


I drew the graph and it looks like a trapezoid. I am not sure where to go from here.
 
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Bones said:

Homework Statement


The net force exerted on a particle acts in the positive x direction. Its magnitude increases linearly from zero at x = 0, to 284 N at x = 3.0 m. It remains constant at 284 N from x = 3.0 m to x = 7.0 m and then decreases linearly to zero at point A where x = 9.2 m. Determine the work done to move the particle from x = 0 to x = 9.2 m by graphically determining the area under the F versus x graph.

Homework Equations



The Attempt at a Solution


I drew the graph and it looks like a trapezoid. I am not sure where to go from here.

Welcome to PF.

What is the area then of your trapezoid?

Isn't Work the product of the Force multiplied by the distance over which it acts?
So when Force is a function of distance isn't it:

Work = \int F_{(x)} dx

And this can be found graphically by measuring the area under the graph.
 


Yeah, I figured it out. It just helps sometimes to find a good example for guidelines. Thanks ;)
 
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