Work done by a variable force question

In summary, the conversation is discussing finding the total work done on an object as it moves from x = 0 to x = 7, given a graph of force vs displacement and the object's mass and initial conditions. The attempt at a solution involved integrating the function and calculating the areas under the positive and negative force regions, resulting in a total work of -0.5J. There is a question about whether this violates the work-energy theorem, but it is explained that the theorem is not violated as the object has some final velocity. There is also a clarification about the sign of the work done.
  • #1
Catalytical
5
0

Homework Statement



246kgu9.jpg


That's a graph of Force in Newtons (y axis) vs displacement in meters (x axis) for an object of mass 3.0kg that is moving along the x-axis and initially starts at rest. I am being asked to find the total work done on the object as it moves from x = 0 to x = 7.


Homework Equations



Net Work = ∫F dx = ΔK
K = 1/2mv^2

The Attempt at a Solution



So, I integrated the function and ended up with a total work of -.5J. What I want to know is if this violates the work-energy theorem or not. If the net work is equal to change in kinetic energy, and the object starts at rest, this would mean that -.5 = 1/2mv^2. Is there something really obvious that I'm missing here?
 
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  • #2
work done (energy) = area under F against x graph (this is another way of seeing ΔK = ∫F.dx
There are 2 clear areas to calculate... when F is +ve and when F is -ve
 
  • #3
By calculating the areas I got total work = -0.5J
 
  • #4
So, I integrated the function and ended up with a total work of -.5J. What I want to know is if this violates the work-energy theorem or not. If the net work is equal to change in kinetic energy, and the object starts at rest, this would mean that -.5 = 1/2mv^2. Is there something really obvious that I'm missing here?

Can you explain why you think the work-energy theorem is violated?

[itex]0.5=\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}[/itex]

If the object started at rest, the last term is 0 and you get

[itex]0.5=\frac{1}{2}mv^{2}_{f}[/itex]

which just says that the object has some final velocity.
 
  • #5
CanIExplore said:
Can you explain why you think the work-energy theorem is violated?

[itex]0.5=\frac{1}{2}mv^{2}_{f}-\frac{1}{2}mv^{2}_{i}[/itex]

If the object started at rest, the last term is 0 and you get

[itex]0.5=\frac{1}{2}mv^{2}_{f}[/itex]

which just says that the object has some final velocity.

Right, but if the total work done is -.5, wouldn't that mean that
[itex]-0.5=\frac{1}{2}mv^{2}_{f}[/itex]
I'm not sure why you ignored the sign on the work done.
 

Related to Work done by a variable force question

What is work done by a variable force?

Work done by a variable force is the amount of energy transferred to an object by a force that changes in magnitude and/or direction as the object moves. It is a measure of the change in the object's kinetic energy.

How is work done by a variable force calculated?

The formula for calculating work done by a variable force is W = ∫F(x)dx, where F(x) is the variable force and dx is the infinitesimal displacement of the object in the direction of the force.

What is the difference between work done by a constant force and a variable force?

The main difference is that a constant force remains the same in magnitude and direction, while a variable force changes in magnitude and/or direction as the object moves. This means that the work done by a constant force is a simple multiplication of the force and displacement, while the work done by a variable force requires integration of the force over the displacement.

Can work done by a variable force be negative?

Yes, work done by a variable force can be negative. This occurs when the force and displacement are in opposite directions, meaning that the force is doing negative work on the object, taking away energy from the object.

What are some real-life examples of work done by a variable force?

Examples of work done by a variable force include pulling a spring, pushing a cart up a hill, and swinging a pendulum. In all of these cases, the force changes in magnitude and/or direction as the object moves, resulting in work done by the force on the object.

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