Introduction

WORK

It is defined as dot product of force and displacement

W = $\vec{F}.\vec{s}$ = F.s.cos$\theta$

where $\vec{F}$ is force acting on a point and $\vec{s}$ is displacement, $\theta$ is the angle between force and displacement.

W = ${\int }_{}^{}\vec{F}.d\vec{x}$

Illustration 1:

A block of ice is drawn through a distance 500 cm along a smooth horizontal surface. The pull in the rope in 100 dyne and the angle between the rope and ground in 30°. Find the work done.

Solution :

Work done = component of the force parallel to the displacement x displacement

= F cos $\theta$ x s = 100 cos 30° x 500 = 43300 erg.

q Since displacement depends on reference frame, hence work also depends on the reference frame.

POWER

The rate at which work is done (by an agent) is known as power, which is delivered to the object by the agent.

In other words,

Power, P = $\frac{dW}{dt}$ = $\frac{\vec{F}.d\vec{x}}{dt}$,

since dW = $\vec{F}.d\vec{x}$

where $\vec{F}$ is the force and d $\vec{x}$ is the displacement in time dt.

= $\vec{F}.\vec{v}$ ($\vec{v}$ =$\frac{d\vec{x}}{dt}$ = the instantaneous velocity of the body)

If we need to calculate average power then

P_av = $\frac{Networkdone}{Timetaken}=\frac{\underset{o}{\overset{T}{\int }}P.dt}{T}$

Illustration 2.

An engine of mass 20 tons pulls a train of 20 wagons, each of mass 20 tons with constant velocity of 72 km/hr on a level track. If the coefficient of kinetic friction is ${\mu }_k$ = 0.01, find the power developed by the engine.

Solution:

The total forward force F exerted by the engine equals the frictional force ( = ${\mu }_k$ mg) on the whole train (including the engine), as the train is moving forward with constant velocity.

The net power developed by the engine = F. v = ${\mu }_k$ (mg)v

= 0.01 x (420 x 10³) x 10 x $\frac{72\times {10}^3}{3600}$= 420 kW.

ENERGY

Energy is the capacity (of a body or agent) to do work.

Mechanical energy is of two types:

Kinetic energy and Potential energy.

KINETIC ENERGY

It is the energy possessed by a body by virtue of its motion. If mass of body is m and speed v the kinetic energy

K.E. = $\frac{1}{2}m{v}^2$

POTENTIAL ENERGY

Potential energy of a body or system is the energy possessed by the body by virtue of its position or strain.

Potential energy function can only be defined for conservative force field only. In case of uniform gravity, potential energy is mgh where m is mass of body and h is height above the chosen reference level.

Illustration 3.

The potential energy of a system of two particles is given by U(x) = a/x² - b/x. Find the minimum potential energy of the system, where x is the distance of separation; a, b are constants.

Solution:

U(x) = $\frac{a}{x^2}-\frac{b}{x}$

$\Rightarrow$ F = $\frac{-dU(x)}{dx}$ $\Rightarrow$ F = -$\frac{d}{dx}\left[\frac{a}{x^2}-\frac{b}{x}\right]$

$\Rightarrow$ F = $\left[\frac{-2a}{x^3}+\frac{b}{x^2}\right]=\left(\frac{a}{x^3}-\frac{b}{x^2}\right)$

When the particle is in equilibrium

F = 0

$\Rightarrow$ 2/x³ - $\frac{1}{x^2}$ = 0

$\Rightarrow$ x = $\frac{2b}{a}$

Therefore, the minimum potential energy of the system is obtained by putting

x = $\frac{2b}{a}$ in U(x) = $\frac{a}{x^2}-\frac{b}{x}$,

U_min $=\frac{a}{{\left(\frac{2b}{a}\right)}^2}-\frac{b}{\frac{2b}{a}}$= $\frac{a^3}{4{\mathbf{b}}^2}-\frac{a}{2}$ = $\frac{a}{2}\left[\frac{a^2}{b^2}-1\right]$.

UNITS AND DIMENSIONS OF WORK, POWER AND ENERGY

Work and Energy are measured in the same units. Power, being the rate at which work is done, is measured in a different unit.

Conversions between Different Systems of Units

1 Joule = 1 Newton x 1 m = 10⁵ dyne x 10² cm = 10⁷ erg

1 watt = 1 Joule/ sec = 10⁷ erg/sec.

1 ft. poundal = 1 poundal x 1 ft = 13825 dyne x 30.48 cm = 4.214 x 10⁵ erg

1kg-m = 1kg- wt x 1 m

= "g" Newton x 1m = "g" Joule = 9.8 Joule

1ft-lb = "g" poundal x 1 ft

32.2 poundal x 1ft

32.2 ft-poundal

1 kwh = 10³ watt x 1 hr = 10³ watt x 3600 sec

= 3.6 x 10⁶ Joule

1HP = 550 ft. $\ell$b/sec (by definition)

= 32.2 x 550 x 4.214 x 10⁵ erg/sec. = 746 x 10⁷ erg/sec.

= 746 watt.

1 MW = 10⁶ watt.

1 cal = 1 calorie = 4.2 Joule

1eV = "e" Joule = 1.6 x 10^-19 Joule

(e = magnitude of charge on the electron in colombs)

POTENTIAL ENERGY OF SPRING

Since spring force is conservative, hence potential energy function can be defined.

dU = –$\vec{F}.d\vec{x}$

$\therefore$ U = ${\int }_{}^{}dU$ = –${\int }_{}^{}\vec{F}.d\vec{x}$

Since F_s = – kx

$\therefore$ ${\int }_0^{u}dU$ =${\int }_0^{x}kx.dx$

U = $\frac{1}{2}k{x}^2$

This is the potential energy stored in the spring of force constant k.

Illustration 4.

Find the maximum energy stored in the spring shown in the figure, for which the block remains stationary on the rough horizontal surface.

Solution: Let the spring be compressed by x. The potential energy stored in the spring

= U = 1/2 Kx² for maximum potential energy in the spring.

N – Mg = 0 $\Rightarrow$ N = mg

Therefore, f_max = mmg

kx_max = mmg $\Rightarrow$ x_max = $\frac{\mu mg}{k}$

The maximum potential energy stored in the spring

= 1/2 kx_max² = $\frac{1}{2}k{\left(\frac{\mu mg}{k}\right)}^2$

= $\frac{{\mu }^2{m}^2{g}^2}{2k}$.