13.5 Work by gravity  (Page 110/3)

Gravity is common to all motions. The most striking aspect of work by gravity stems from the fact that the force due to gravity is constant. As such, value of work is constant for a given displacement irrespective of the state of motion of the particle (i.e velocity of particle).

We consider two different situations for the analysis of work :

• Particle is subjected to single force due to gravity.
• Particle is subjected to force due to gravity and other forces.

Two situations are different in their manifestations. For example, projecting an object in vertical direction with initial velocity is different to a situation in which the object (say a lift) is pulled up by the rope attached to it. In the first case, only force that influences the motion of the object is the force due to gravity, whereas the motion of the object in the second case is determined by the force of gravity and tension in the string pulling the object. The situation, in the first case, is described by initial velocity. On the other hand, we will need force analysis of the forces with a free body diagram to know the forces and resulting acceleration. These two situations are shown in the figure.

There are many such situations in which force(s) other than garvity plays a role in the motion of an object. When we lift a book from the table, two forces act on the book (i) weight of the book and (ii) force applied by the hand. Note that gravity is always present and is common to all situations.

Particle is subjected to single force due to gravity.

We discuss here two different profile of motions (i) vertical and (ii) inclined motion :

Vertical motion

When we throw a particle like object with an initial velocity vertically upwards, the force due gravity decelerates the motion by doing a negative work on the object. For a vertical displacement, y, the work done is :

$$\begin{array}{l}{W}_{\mathrm{G(up)}}=(-mg)r=-mgy\end{array}$$

Kinetic energy of the particle is correspondingly reduced by the amount of work :

$$\begin{array}{l}{K}_{\mathrm{f(up)}}=K_{\mathrm{i(up)}}+W_{\mathrm{G(up)}}=K_{\mathrm{i(up)}}-mgy\end{array}$$

It is clear from "work-kinetic energy" theorem that initial kinetic energy is equal to "mgh", where "h" is the maximum height.

$$\begin{array}{l}{K}_{\mathrm{f(up)}}=K_{\mathrm{i(up)}}+W_{\mathrm{G(up)}}=mgh-mgy=mg(h-y)\end{array}$$

During this upward motion, force due to gravity transfers energy "from" the particle. This process continues till the particle reaches the maximum height, when kinetic energy is reduced to zero (y = h):

$$\begin{array}{l}{K}_{\mathrm{f(up)}}=0\end{array}$$

A reverse of this description of motion takes place in downward motion. The work by gravity is positive in the downward motion. During this part of motion, the kinetic energy of the particle increases by the amount of work :

$$\begin{array}{l}{W}_{\mathrm{G(down)}}=mgr=mg(h-y)\end{array}$$

where “h” is the maximum height attained by the particle. During this downward motion, force due to gravity transfers energy "to" the particle. As a consequence, kinetic energy increases, which is given by :