# Work with varying forces. Work energy theory

• EEintraining
In summary, the net force along the x-axis that has x-component F(x)= -12.0N + (0.300 N/m^2)x^2 is applied to a 3.70 kg object that is initially at the origin and moving in the -x-direction with a speed of 8.80 m/s . This force is causing the object to move with a speed of 8.80 m/s at the point x = 7 m.
EEintraining

## Homework Statement

A net force along the x-axis that has x-component F(x)= -12.0N + (0.300 N/m^2)x^2 is applied to a 3.70 kg object that is initially at the origin and moving in the -x-direction with a speed of 8.80 m/s .

## Homework Equations

w= f * s

w= ∫F(x)dx (varying forces)

w = Δk

k = 1/2 mv^2

## The Attempt at a Solution

I am stuck on how to approach this question. I have seen many of these types of questions answered with potential and kinetic energy, but we have not covered that in class so far. We have only covered kinetic energy and the work energy theory. I realize that the force is varying so that It will eventually turn the object around in the positive x direction so that I can calculate the speed at 7m, but I am not sure what to do with the information that I have. Any help or pointers to help me see what I am missing would be appreciated.

Thanks

EEintraining said:

## Homework Statement

A net force along the x-axis that has x-component F(x)= -12.0N + (0.300 N/m^2)x^2 is applied to a 3.70 kg object that is initially at the origin and moving in the -x-direction with a speed of 8.80 m/s .

## Homework Equations

w= f * s

w= ∫F(x)dx (varying forces)

w = Δk

k = 1/2 mv^2

## The Attempt at a Solution

I am stuck on how to approach this question. I have seen many of these types of questions answered with potential and kinetic energy, but we have not covered that in class so far. We have only covered kinetic energy and the work energy theory. I realize that the force is varying so that It will eventually turn the object around in the positive x direction so that I can calculate the speed at 7m, but I am not sure what to do with the information that I have. Any help or pointers to help me see what I am missing would be appreciated.

Thanks
What's the question you're trying to answer ?

If you need to get the speed at 7 m then you need to use the fact that a = dv/dt to get the relationship for speed. (you will need to use the chain rule to change dv/dt to something else involving dv/dx).

Yes that part is important.

What is the speed of the object with it reaches the point x = 7 m?

Rock.freak i am not sure how to get the acceleration. It starts out going -8.8 m/s, so its kinetic energy is 1/2*m*v^2 or 143.264 J. So I need to apply 143.264 J just to stop it, then the force function will move it in the positive direction. So if I integrate the force function, can I set the 143 J equal to that function, solve for x, and then find the distance between it and +7m?

Does that make any sense?

Use the work-energy theorem.

If all you need is the velocity at x = 7m, then there's no need for you to find the acceleration.

write f as ma and a as dv/dt. multiply both sides by dx. then it becomes ..vdv=...dx
integrate both sides

altamashghazi said:
write f as ma and a as dv/dt. multiply both sides by dx. then it becomes ..vdv=...dx
integrate both sides
The result of doing this is the Work-Energy theorem for this particular force.

## 1. What is work energy theory?

The work energy theory is a concept in physics that describes the relationship between work and energy. It states that the work done on an object is equal to the change in its kinetic energy.

## 2. How do varying forces affect work?

Varying forces can affect the work done on an object by changing the magnitude and direction of the force applied. This can result in either an increase or decrease in the amount of work done.

## 3. What is the formula for calculating work with varying forces?

The formula for calculating work with varying forces is W = ∫F(x)dx, where W is the work done, F(x) is the varying force, and dx is the displacement of the object.

## 4. How does work energy theory relate to real-life situations?

Work energy theory can be applied to real-life situations, such as lifting a heavy object or pushing a car. In these cases, the work done on the object is equal to the change in its kinetic energy, which can be observed through its movement.

## 5. What are some practical applications of work energy theory?

Work energy theory has many practical applications, such as in engineering, construction, and transportation. It is also used in sports and exercise science to understand the relationship between work and energy in the human body.

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