[Physics Class Notes] on Work Done By A Variable Force Pdf for Exam

Applying force on an object causes that object to move in the direction of the force. This movement in relation to the force is defined as work. You can complete work using a constant force or a variable force. 

What is a Constant Force?

Work done by a force can be divided into work from a constant force and work from a variable force. In the former kind, the magnitude and direction of the force remain unaltered. In such a case, work (W) is equal to the force applied (F) multiplied by displacement ([Delta]x). Therefore,

W = F x [Delta]x

What is Variable Force?

Work done by the variable force is a bit more complex. In such a case, the magnitude and direction of force can change at any time during the work. Most of the work that we complete in our daily life is an example of variable force work. Calculating the same is quite complex and requires integration.

To form a better understanding of the same, let us consider the workings of a spring.

Revising Hooke’s Law

Hooke’s Law states that the spring force for a compressed or stretched spring is equal in magnitude to the force for extension or compression of the spring. However, this spring force has an opposite direction to this extension. 

The figure above relates the force on the spring vs displacement when displacement is 0 for an unstretched spring. Us is the elastic potential energy for a stretched spring. This force-displacement graph for spring can help in assessing force according to Hooke’s Law.

Fs = -kx

Thus, Ws = Fs x vdt

Integration and Formula for Variable Force Work

For work done by a variable force, however, you need to apply integration to arrive at accurate results. Therefore,

[W_{s} = int_{0}^{t}F_s .vdt]

[W_{s} = int_{0}^{t} – kx v_{x} dt]

[W_{s} = int_{x^0}^{x} – kx dx]

[W_{s} = -frac{1}{2} kDelta x^{2}]

Consequently, by using this approach mentioned above, one can easily derive the work done by variable force.

Quick Exercise – 1

Mass (m) of an object is 2kg. This object undergoes variable force in direction ‘x’. Force variation is a function, Fx = (3 + .2x) N. Determine the work done when object moves from x = 0 to x = 5.

Solution

Work done [ = int_{x^a}^{x^b} fdx]

[W = int_{0}^{5} (3+.2x) dx]

W = 17.5 J

Quick Exercise – 2

A bullet weighing 20g is moving at a velocity of 500m/s. This bullet strikes a windowpane and passes through it. Now, its velocity is 400m/s. Calculate work done by a bullet when passing through this obstacle.

Solution

We need to determine the change in kinetic energy in this equation. You know that kinetic energy change [Delta] (K.E.) =[frac{1}{2} * left ( mv_{1}^{2} – mv _{2}^{2}right)]

Therefore, in this equation, m is 20g or 0.02 kg. Initial velocity (v1) is 500m/s, while final velocity (v2) is 400m/s.

Thus, [Delta] (K.E.) = [frac{1}{2} * {0.02 (500)^{2} – 0.02 (400)^{2}}]

[Delta] (K.E.) = 900 Joules.

Deriving Work Done by a Constant Force with Integration

Just as you can derive the work for a variable force using calculus, you can do the same for work done by a constant force. In such a calculation, pressure remains unchanged, which is why you can take it out of the equation immediately. 

After doing this, you will arrive at an equation, where

Work = [Pint_{a}^{b} dv].

[W = PDelta V]

As you can see, the product is the same that we would have evaluated from considering force and distance. Thus, following this integration method for work done by a constant force is redundant.

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This article provides an explanation of work done by a variable force, its formula, and derivations. The frequently asked questions at the end of this article can help you with your doubts if you encounter any while understanding the same.

Variable Force

When a force acts upon a body changes with time(varies) it is known as variable force.  Work done by a variable force is calculated by  Integration.

Formula

Integration can be used for calculating both, that is, work done by a variable force and work done by a constant force.

[W_{s} = int_{0}^{t}Fs .vdt]

[W_{s} = int_{0}^{t} – kx v_{x} dt]

[W_{s} = int_{x0}^{x} – kx dx]

[W_{s} = -frac{1}{2} kDelta x^{2}]

Derivation

A force does work when it results in movement.

Work done by a constant force of magnitude F on a point moves a displacement

[ Delta x]x  in the direction of the force is the product

[W=F ⋅ Delta x]

In the case of a variable force, integration is essential to calculate the work done. For example, let us consider work done by a spring. Hooke’s law states that the spring or restoring force of a perfectly elastic spring is proportional to its extension but opposite to the direction of extension.

All the small contributions to the work done during small time intervals, for a variable force.