Work Done By A Variable Force

What is Work?

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 (Δx). Therefore,

W = F x Δx

What is Variable Force?

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

Ws = ∫t0Fs ⋅ vdt

Ws = ∫t0 – kx vxdt

Ws =∫xxo – kx dx

Ws = -1/2kΔx2 

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 = xbxa fdx

W = 50 (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 Δ (K.E.) = ½ x (mv12 – mv22)

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, Δ (K.E.) = ½ x {0.02 (500)2 – 0.02 (400)2}

Δ (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 = P∫ba dV.

W = PΔV

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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FAQ (Frequently Asked Questions)

1. What is the formula for deriving work from a constant force?

You can multiply constant force with the displacement occurring due to the application of that force to assess the exact work done by a constant force.

2. What are the two requirements for work done?

An external force must be applied on an object due to which, the object must move in the direction of the applicable force.

3. What is the SI unit of work?

The SI unit for measuring work is Joule.