Work Done by Force F: bx^3 over \Deltax=2.6m

In summary, a force of F = bx^{3} with a value of b = 3.7 N/m3 is applied in the x-direction and moves an object from x = 0.0 m to x = 2.6 m. The relevant equation for work is dW = F_xdx and to find the total work, this expression must be integrated.
  • #1
JasonPF
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Homework Statement



A force F = bx[tex]^{3}[/tex] acts in the x-direction. How much work is done by this force in moving an object
from x = 0.0 m to x = 2.6 m? The value of b is 3.7 N/m3.

[tex]\Delta[/tex]x = 2.6m


Homework Equations


W = F[tex]\Delta[/tex]x
F = bx[tex]^{3}[/tex]


The Attempt at a Solution


I have no attempted solutions to this problem. I feel as if I simply am overlooking something very simple. I'm not necessarily looking for an answer, but perhaps someone could point out a concept I may have missed or not shown here that would spark my brain into solving it.

Any help is greatly appreciated.
 
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  • #2
You have the wrong relevant equation for work. When the force doing work is variable, work is not just "force times displacement." The correct equation for an element of work in one dimension is

[tex]dW=F_xdx[/tex]

To get the total work you must integrate this expression.
 
  • #3
Thank you very much, as I thought, I was just over looking something. That solved my problem.
 

FAQ: Work Done by Force F: bx^3 over \Deltax=2.6m

1. What does "Work Done by Force F: bx^3 over \Deltax=2.6m" mean?

The statement is describing the work done by a force, represented as F, on an object along a distance of 2.6 meters, where the force varies with the position of the object and is given by the equation bx^3.

2. How do you calculate the work done in this situation?

To calculate the work done, you would need to integrate the force function (bx^3) with respect to the position variable (x) over the distance of 2.6 meters. This will give you the total amount of work done by the force on the object.

3. What is the unit of measurement for work done?

The unit of measurement for work done is joules (J), which is equivalent to a force of one newton applied over a distance of one meter.

4. Can you give an example of a real-life scenario where this equation would be applicable?

This equation could be applicable in situations where a varying force is applied to an object over a specific distance. For example, calculating the work done by an engine on a car as it accelerates from rest to a certain speed over a distance of 2.6 meters.

5. What other factors might affect the work done by this force?

The work done by this force may also be affected by other factors such as the mass of the object, the angle at which the force is applied, and the presence of other forces acting on the object.

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