# Calculation of work done by a variable force

• navneet9431
In summary, the author discusses the calculation of work done by variable forces by analyzing a small fraction of displacement and making assumptions about the angle and direction of the force. The questions raised include how to integrate when the angle is continuously changing and how to determine the direction of the force. The correctness of these assumptions is also discussed, with the conclusion that the magnitude and direction of the force are both variables and must be included in the integrand. It is also noted that there could be other forces acting on the object and that the author assumes prior knowledge of the force's magnitude and direction.
navneet9431
Gold Member

## Homework Statement

See the explanation of work done by variable forces given in my textbook.

For calculation of work done by variable forces the author analyses only a small fraction of displacement "ds".
And then he makes **assumption** that the force is acting on the particle at an angle theta and that too in the *upward* direction.And then he integrates them.
My question is how integrating them would give us the total work done by the variable force.Because at different positions the angle at which the force is acting would change.It would not be 'theta' for all instants.
And my next question is that how should we determine in which direction the force is acting?
Like in my book the author assumed that it would act like this(in the upward direction),

So what if I say that the force would not act like that but would act like this in the downward direction?

W=Fcos(theta)*S

## The Attempt at a Solution

I think that the direction of force should be determined by the curve if the path.If the path bends inwards then the force would act downwards and if the path is bending upwards then the force would act upwards.
Is my thinking correct?
I will be thankful for any help!

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navneet9431 said:
at different positions the angle at which the force is acting would change
The author allows that θ is a variable.

Ok then integrating them would be wrong as the angle would be changing continuously at every instant.
So integrating them would not give us the actual work done,but slightly different value.
Am I correct?
haruspex said:
The author allows that θ is a variable.

navneet9431 said:
then integrating them would be wrong
No, integrating is fine as long as it is treated as a variable. The mistake would be to move the cos(θ) term outside of the integral.

navneet9431
haruspex said:
No, integrating is fine as long as it is treated as a variable. The mistake would be to move the cos(θ) term outside of the integral.
Thanks!
Now check my this reasoning,is it correct?

"I think that the direction of force should be determined by the curve of the path.If the path bends inwards then the force would act downwards and if the path is bending upwards then the force would act upwards."

navneet9431 said:
the direction of force should be determined by the curve if [of] the path
The author is not assuming the force is constant either, neither in magnitude nor direction.

navneet9431 said:
"I think that the direction of force should be determined by the curve of the path.If the path bends inwards then the force would act downwards and if the path is bending upwards then the force would act upwards."
You are assuming that this force is the only force acting on the object. This is not necessarily true.

A representative diagram showing a force in a direction does not mean that the force is always in that direction, or has that magnitude. In general, both the magnitude and direction (θ) are variables and are part of the integrand. Note also that ds itself changes as you follow the path. Also note post #7 above, There could be three different forces acting on the object in different directions, and you may need to calculate the work done by one of the forces.

The author is assuming you already know (in advance) the magnitude and direction of the force along the path, in terms of their variation with path position s.

## 1. What is work done by a variable force?

The work done by a variable force is the product of the magnitude of the force and the displacement in the direction of the force. It is a measure of the energy transferred to an object by the force.

## 2. How is work calculated for a variable force?

To calculate work done by a variable force, you need to integrate the force function with respect to displacement. This gives you the area under the force-displacement curve and represents the work done.

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

The main difference is that a variable force changes in magnitude or direction, while a constant force remains the same. This results in a more complex calculation for work done by a variable force, as it requires integration.

## 4. 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, resulting in a negative value for work. It represents energy being taken away from an object rather than being added to it.

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

Examples of work done by a variable force include pushing a shopping cart, rowing a boat, and swinging a pendulum. In all of these situations, the force applied changes in magnitude and/or direction, resulting in a variable force.

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