Lesson 3 of 13
In Progress

Notions of Work and Kinetic Energy: The Work-Energy Theorem

The following relation for rectilinear motion under constant acceleration a has been encountered in previous lesson.

v2 u2 = 2as where u and v are the initial and final speeds and s the distance traversed. Multiplying both sides by m/2, we have

1/2 mv2 -1/2 mu2 = mas = Fs …………………………………………………………………….(2a)

where the last step follows from Newton’s Second Law. We can generalise Eq. 1 to three dimensions by employing vectors

v2 u2 = 2 a.d

Once again multiplying both sides by m/2, we obtain

1/2 mv2 -1/2 mu2 = ma.d = F.d ………………………………………………………………..(2b)

The above equation provides a motivation for definitions of work and kinetic energy. The left side of the equation is the difference in the quantity ‘half the mass times the square of the speed’ from its initial value to its final value. We call each of these quantities the ‘kinetic energy’, denoted by K . The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by W. Eq. (2) is then

Kf – Ki = W ……………………………………………………………………………………….(3)

where Kf and Ki are respectively the final and initial kinetic energies of the object. Work refers to the force and displacement over which it acts. Work is done by a force on the body over a displacement.

Equation (2) is also a special case of the work-energy (WE) theorem: The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to varying force in a later section.Example 2
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.oo km. It hits the ground with a speed of 50.0 m s1. What is the work done by the unknown resistive force?

Answer

(a) The change in kinetic energy of the drop is

ΔK = 1/2 mv2 0

= 1/2 × 103× 50 × 50

= 1.25 J where we have assumed that the drop is initially at rest.

Assuming that g is a constant with a value 10 m/s2 , the work done by the gravitational force is,

Wg = mgh

= 103× 10 × 103

= 10.0 J

(b) From the work-energy theorem

ΔK = Wg + Wr where Wr is the work done by the resistive force on the rain drop. Thus

Wr = ΔK – Wg

= 1.25 – 10

=-8.75 J is negative