Work done by a constant force is equal to the product of the force and displacement of the object in the direction of force.

Work done = Force x Displacement of the object in the direction of force

`W = F* s`
or `W = Fs cos theta`

Work is a scalar quantity. Its SI unit is joule and CGS unit is erg.
`1 text(joule) = 10^7 text(erg)`

A force is conservative, if the work done by the force in displacing an object from one point to another point is independent of the path followed by the object and depends only on the end points.

e.g. Gravitational force, electrostatic force and elastic force of a spring are all conservative forces.

If the amount of work done in moving an object against a force from one point. to another point depends on the path along which the body moves, then such a force is called a non-conservative force. The work done in moving an object against a non-conservative force along a closed path is not zero.

e.g. Forces of friction and viscosity are non-conservative forces.

Energy is defined as the capacity or ability of a body to do work. Energy is scalar quantity and its units and dimensions are the same as that of work. Thus, SI unit of energy is joule. There are so many types of energy e.g. kinetic, potential, electrostatic, magnetic, geothermal, elastic, solar etc. Some of them are described below. Some other commonly used units of energy are

`1 erg = 10^-7` J
`1 cal = 4.186 J ~~ 4.2 J`
`1 kcal = 4186 J, 1 kWh = 3.6 xx 10^6 J`

and 1 electron volt `= 1 eV = 1.60 xx 10^-19 J`

`text(Kinetic Energy)`
Kinetic Energy (KE) is the capacity of a body to do work by virtue of its motion. Motion may be either transnational or rotational.

A body of mass m, moving with a velocity v, has a kinetic energy
`K =1/2 mv^2`
Thus , `K prop m`
`K prop v^2`

Kinetic energy of a body is always positive irrespective of the sign of velocity v. Negative kinetic energy is impossible.

Kinetic energy is correlated with momentum as
`K = p^2/(2m) ` or `p = sqrt( 2mK)`

Kinetic energy for a system of particle will be
`K =1/2 sum_i m_i v_i^2`

Kinetic energy depends on the frame of reference. Kinetic energy of a passenger sitting in a running train is zero in the frame of reference of the train but is finite in the frame of reference of the earth.

`text(Potential Energy)`
Potential Energy (PE) is energy of the body by virtue of its position, configuration or state of strain. The relation between potential energy and work done is
`W = - Delta U`
where, `DeltaU` is change in potential energy.

Change in potential energy of a body between any two points is equal to the negative of work done by the conservative force in displacing the body between these two points, without there being any change in kinetic energy. Thus,
`dU = - dW = - F * dr`
and `U_2 - U_1 = - W`
`= -int_(r_1)^(r_2) F * dr`

Value of the potential energy in a given position can be defined only by assigning some arbitrary value to the reference point. Generally, reference point is taken at infinity and potential energy at infinity is taken as zero. In that case,
`U = - W = - int_(oo)^r F * dr`

Potential energy is a scalar quantity but has a sign. It may be positive as well as negative.

► Generally potential energy is of three types

`text(Gravitational Potential Energy)`
It is the energy associated with the state of separation between two bodies which interact via the gravitational force. The gravitational potential energy of two particles of masses `m_1` and `m_2` separated by a distance r is
`U = (-Gm_1m_2)/r`

Generally, one of the two bodies is our earth of mass M and radius R If m is the mass of the other body, situated at a distance r(`r >= R`) from the centre of earth, the potential energy of the body
`U(r) = -(GMm)/r`

`text(Some Important Points)`
If a body of mass m is raised to a height h from the surface of earth, the change in potential energy of the system
(earth + body) comes out to be
`DeltaU=(mgh)/(1+h/R)` or `DeltaU~~mgh`, if `h< Thus, the potential energy of a body at height h,
i.e. mgh is really the change in potential energy of the system for `h < < R`.

For the gravitational potential energy, the zero of the potential energy is chosen to be the ground.

`text(Elastic Potential Energy)`
Whenever an elastic body (say a spring) is either stretched or compressed, work is being done against the elastic spring force. The work done is `W = 1/2 kx^2`
where, k is spring constant and x is the displacement
and elastic potential energy `U = 1/2kx^2`
Elastic potential energy is always positive.

`text(Electric Potential Energy)`
The electric potential energy of two point charges `q_1` and `q_2` separated by a distance r in vacuum is given by
`U = 1/(4 pi epsi_0) (q_1q_2)/r`
where , `1/(4 piepsi_0) = 9.1 xx 10^9 N-m^2//C^2` = constant

Elastic potential energy is equal to work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched. From Hooke's law, the force required to stretch the spring will be directly proportional to the amount of stretch.

`F = - kx` [force by spring]

Then, the work done to stretch the spring by a distance x is
`W = PE = 1/2kx^2` [work done by stretcher]

Since the change in potential energy of an object between two positions is equal to the work that must be done to move the object from one point to the other, the calculation of potential energy is equivalent to calculating work. Since, the force required to stretch a spring changes with distance, the calculation of the work involves an integral.
`W = int_0^x kx dx = k x^2/2`

Work can also be visualized as the area under the force curve. Here, object is displaced slowly.
If the potential energy function U is known, the force at any point can be obtained by taking the derivative of the potential.

`F_y = -(dU)/(dy) = (-d)/(dy) ( mgy)`
`F_x = - (dU)/(dx) = -d/(dx) ( 1/2 kx^2)`
`F_y = -mg , F_x = -kx`

The rate of doing work or the rate at which energy is transferred or used or transformed is called power.
If work W is done in time t, then Power ,
`P = text( work )/text( time )`
` => P = W/t = (F * s)/t = F * v cos theta`

The SI unit of power is watt in honour of James Watt having the symbol W. We express larger rate of energy transfer in kilowatt (kW).
` 1 W = 1 J s^-1`
or `1 kW = 1000 W = 1000 Js^-1`
`=> 1 MW = 10^6 W`
`1 HP text((horse power)) = 746 text((watt)) W`

`text(Commercial Unit of Energy)`
The commercial unit of electric energy is kilowatt-hour (kWh).
It is the amount of electric energy consumed by an appliance of power 100 W in one hour.

`1 kWh = 1 k w xx 1 h = 1000 W x 3600 s`
`= 1000 Js^-1 xx 3600 s`
`1 kWh = 3.6 xx 10^6 J`

A collision is said to occur between two objects, either if they are physically collide against each other or if the path of one object is affected by the force exerted by the other object.

► There are mainly two types of collision

1. `text(Elastic collision)` If there is no loss of kinetic energy during a collision, then it is called an elastic collision.

Characteristics of elastic collisions are as follows
(i) The kinetic energy is conserved.
(ii) The momentum is conserved.
(iii) Total energy is conserved.
(iv) Forces involved during the collision are conservative.

2. `text(Inelastic collision)` If there is a loss of kinetic energy during a collision, then it is called an inelastic collision.

Characteristics of inelastic collisions are as follows
(i) The kinetic energy is not conserved.
(ii) The momentum is conserved.
(iii) Total energy is conserved.
(iv) Some or all of the forces involved are non-conservative.

If particles of masses `m_1` and `m_2` move with velocities `u_1` and `u_2` before collision and after collision with velocities `v_1` and `v_2`, then from law of conservation of momentum,

`m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2`

If an object of mass m strikes horizontally a wall with velocity v and bounces back with the same velocity, then the change in momentum `= 2mv`.

Consider two perfectly elastic objects A and B of masses `m_1` and `m_2` moving along the same straight line with velocities `u_1` and `u_2` respectively.

Let `u_1 > u_ 2` after some time, the two objects collide head-on and continue moving in the same direction with velocities `v_1` and `v_2` , respectively. The two objects will separate after the collision, if `v_2 > v_1`

As linear momentum is conserved inelastic collision, so

`m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2` .........................(i)

Since, kinetic energy is also conserved in an elastic collision, so

`1/2m u_1^2 + 1/2m_2u_2^2 = 1/2 m_1v_1^2 + 1/2m_2v_2^2` ......................(ii)

After solving the above two equations, we get

`v_1 = ((m_1 -m_2 )/ ( m_1 + m_2) ) u_1 + ((2m_2)/(m_1 + m_2 )) u_2`

`v_2 = (( m_2 - m_1)/ ( m_1 + m_2 )) u_2 + ( ( 2 m_1)/ ( m_1 + m_2)) u_1`