# 13.5 Work by gravity  (Page 2/3)

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$\begin{array}{l}{K}_{\mathrm{f\left(down\right)}}={K}_{\mathrm{i\left(down\right)}}+{W}_{\mathrm{G\left(up\right)}}={K}_{\mathrm{i\left(down\right)}}+mg\left(h-y\right)=0+mg\left(h-y\right)=mg\left(h-y\right)\end{array}$

The process continues till the particle returns to the initial position. Its velocity increases and becomes equal to the velocity of projection on its return to the initial position. For y = 0,

$\begin{array}{l}{K}_{\mathrm{f\left(down\right)}}==mgh\end{array}$

As such, kinetic energy, at the end of round trip, is equal to that in the beginning.

$\begin{array}{l}⇒{K}_{\mathrm{f\left(down\right)}}={K}_{\mathrm{i\left(up\right)}}\end{array}$

For the round trip, net work by gravity is zero. Net transfer of energy “to” or “from” the particle is zero. Initial kinetic energy of the particle is retained at the end of round trip.

Important aspect of this description of motion under gravity is that the end results are specific to gravity. The same result will not hold for motion, which is intervened by force like friction. We shall know that the difference arises due to the difference in the nature of two forces. We shall discuss these difference in detail in a separate module on conservative force.

## Motion along incline

We shall, now, consider projection of a block on a smooth incline with initial velocity "v". Only force acting on the block is force due to gravity. The component of force is mg sinθ, which opposes motion in upward direction and aids motion in downward direction. Work done for going up the incline for a displacement “y” is :

$\begin{array}{l}{W}_{\mathrm{G\left(up\right)}}=\left(-mg\mathrm{sin}\theta \right)xr\end{array}$

where "r" is any intermediate length along the incline covered by the block. Considering trigonometric ratio of sine of the angle of incline, we have :

$\begin{array}{l}⇒y=r\mathrm{sin}\theta \end{array}$

Hence,

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

This is an important result as it underlines that the work depends only on the vertical component of displacement i.e. the displacement along the direction in which force is applied - not on the component of displacement in the direction perpendicular to that of force. Work by gravity for horizontal displacement is zero as force and displacement are at right angle. This independence of work by gravity from horizontal displacement is further reinforced with the concept of conservative force and corresponding concept of potential energy as discussed in the corresponding modules.

The initial kinetic energy of the block decreases as force due to gravity transfers energy from the block :

$\begin{array}{l}{K}_{\mathrm{f\left(up\right)}}={K}_{\mathrm{i\left(up\right)}}-mgy\end{array}$

This process continues till the particle reaches the maximum height, when kinetic energy is reduced (speed is zero) to zero :

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

A reverse of this description of motion takes place in downward motion. The work by gravity is positive (but same in magnitude) and kinetic increases by the amount of work :

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

where “h” is the maximum height attained by the particle.

The process continues till the particle returns to the initial position, when its velocity is same as that in the beginning. As such kinetic energy at the end of round trip is equal to that in the beginning.

$\begin{array}{l}⇒{K}_{\mathrm{f\left(down\right)}}={K}_{\mathrm{i\left(up\right)}}\end{array}$

For the roundtrip, net work by gravity is zero. Net transfer of energy “to” or “from” the particle is zero. Initial kinetic energy of the particle is retained at the end of round trip.

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