Calculating the work done from an equation for variable force

In summary, the given force equation F = F0(x/x0 - 1) can be used to find the work done by the force in moving a particle from x = 0 to x = 2x0 m. By taking the integral from 0 to 2x0 of (1.5 * ((x/4.9) - 1)), the work done is calculated to be -7.35 J. To find the integral, the equation [(x/a) - 1]dx was used, with a being a constant. The 1.5 was then multiplied at the end.
  • #1

Homework Statement


The force on a particle is directed along an x axis and given by F = F0(x/x0 - 1) where x is in meters and F is in Newtons. If F0 = 1.5 N and x0 = 4.9 m, find the work done by the force in moving the particle from x = 0 to x = 2x0 m.

Homework Equations


F = force, w = work, x = displacement
W=F*x
∫F(x)dx = W

The Attempt at a Solution


Integral from 0 to 9.8 of ( 1.5 * (( x / 4.9) - 1) ) = -7.35 J
 
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  • #2
Hi giveortake. Welcome to PF!

You should show us how you calculated your answer so we can see where you may have gone wrong.

It might be a good idea to do a graph of F(x). What shape is the graph? Where are the x and y intercepts? What does the area between the F(x) and the x-axis represent? The graphs shows two areas, one below and one above the x axis. What can you say about the two areas?

AM
 
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  • #3
That's not what I got.
Show the intermediate step: Integral of [(x/a) - 1]dx. Where a is a constant. The 1.5 can be moved out of the integral and multiplied at the end.
 

What is work and how is it related to force?

Work is the measure of energy transfer that occurs when a force is applied to an object and that object moves in the direction of the force. It is directly related to force, as the amount of work done depends on the magnitude and direction of the force.

What is the equation for calculating work done from a variable force?

The equation for calculating work done from a variable force is W = ∫F(x)dx, where F(x) represents the variable force and dx represents the infinitesimal distance over which the force is applied.

How do you determine the limits of integration for calculating work done from a variable force equation?

The limits of integration can be determined by identifying the initial and final positions of the object, and then integrating the force function over the distance between those two points.

What are the units of work and how are they related to the units of force and distance?

The unit of work is joules (J) in the SI system. It is related to the units of force (newtons, N) and distance (meters, m) by the equation W = Fd, where d is the distance over which the force is applied.

Can the work done from a variable force equation ever be negative?

Yes, the work done can be negative if the force and displacement are in opposite directions. This indicates that the force is acting in the opposite direction of the object's motion, resulting in a decrease in its kinetic energy.

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