Power, P = 100 W
Time of usage, t = 8 hour \(\times\) 30 days = 240 hours
Electrical energy consumed is the product of power and time of usage.
Electrical energy = power \(\times\) time of usage = P \(\times\) t
= 75 watt \(\times\) 240 hour = 18000 watt hour = 18 kilowatt hour = 18 kWh
1 electrical unit = 1 kWh
Electrical energy = 18 unit

Characteristics of elastic collision are
1. Total momentum remains conserved
2. Total kinetic energy remains conserved.
3. In elastic collision conservative forces are involved. Hence total kinetic energy is conserved.
4. In elastic collision, mechanical energy is not dissipated.
Characteristics of inelastic collision are
1. Total momentum is conserved.
2. Total kinetic energy is not conserved.
3. Forces involved are non-conservative forces
4. Mechanical energy is dissipated into heat, light, sound etc.

In Physics, work is said to be done by the force when applied on a body displaces it. To do work, energy is required. But, generally work refers to both physical and mental work. In fact, any activity can be called as work.

(i) For an isolated system, the sum of all forms of energy i.e., the total energy remains the same in any process irrespective of whatever internal changes may take place.
(ii) This means that the energy disappearing in one form reappears in another form. This is known as the law of conservation of energy.

F = k_Ax_A = k_Bx_B
\({ x }_{ A }=\frac { F }{ { k }_{ A } } ,{ x }_{ B }=\frac { F }{ { k }_{ B } } \)
The work done on the springs are stored as potential energy in the springs.
\({ U }_{ A }=\frac { 1 }{ 2 } { k }_{ A }{ x }_{ A }^{ 2 };\quad { U }_{ B }=\frac { 1 }{ 2 } { k }_{ B }{ x }_{ B }^{ 2 }\)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ A }{ x }_{ A }^{ 2 } }{ { k }_{ B }{ x }_{ B }^{ 2 } } =\frac { { { k }_{ A }\left( \frac { F }{ { k }_{ A } } \right) }^{ 2 } }{ { { k }_{ B }\left( \frac { F }{ { k }_{ B } } \right) }^{ 2 } } =\frac { \frac { 1 }{ { k }_{ A } } }{ \frac { 1 }{ { k }_{ B } } } \)
\(\frac { { U }_{ A } }{ { U }_{ B } } =\frac { { k }_{ B } }{ { k }_{ A } } \)
k_A > k_B implies that U_B > U_A.Thus, more work is done on B than A.

S.No.	Conservative forces	Non-Conservative forces
1	Work done is independent of the path	Work done depends upon the path
2	Work done in a round trip is zero	Work done in a round trip is not zero.
3	Total energy remains constant	Energy is dissipated as heat energy
4	Work done is completely recoverable	Work done is not completely recoverable
5	Force is the negative gradient of potential energy	No such relation exists
6	Examples: Elastic spring force, electrostatic force, magnetic force, gravitational force, etc.	Eg: Frictional forces, viscous force

Various types of potential energy are
(i) The energy possessed by the body due to gravitational force gives rise to gravitational potential energy.
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.
U = mgh
(ii) The energy due to spring force and other similar forces give rise to elastic potential energy.
a) At the equilibrium position x = 0 potential energy is \(U=\frac{1}{2} k x^{2}\)
b) If the initial position is not zero and if the mass is changed from position x_i to x_f, if then elastic potential energy is \(U=\frac{1}{2} k\left(x_{f}^{2}-x_{i}^{2}\right)\)
(iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy.
Electrostatic potential energy is the work done to arrange two charges q₁ and q₂ at a separation \(r=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)

Work-Kinetic Energy Theorem
Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.
The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs
The constant force is given by the equation,
F = ma
The third equation of motion can be written as,
\(v^{2} =u^{2}+2 a s  \)
\(a =\frac{v^{2}-u^{2}}{2 s}\)
Substituting for a in equation (2),
\(F=m\left(\frac{v^{2}-u^{2}}{2 s}\right)\)
Substituting equation (2), (1)
\(w=m\left(\frac{v^{2}}{2 s} s\right)-m\left(\frac{u^{2}}{2 s} s\right)  \)
\(w=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
The expression for kinetic energy:
The term \(\left(\frac{1}{2} m v^{2}\right)\) in the above equation is the kinetic energy of the body of mass (m) moving with velocity(v).
\(K E=\frac{1}{2} m v^{2}\)
Kinetic energy of the body is always positive. From equations (4) and (5)
\(\Delta K E =\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}  \)
\(\text {Thus, } W =\Delta K E\)
The expression on the right hand side (RHS) of equation (6) is the change in kinetic energy (\(\Delta\)KE) of the body.
This implies that the work done by the force on the body changes the kinetic energy of the body, This is called work-kinetic energy theorem.
The work-kinetic energy theorem implies the following.
1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

Mass per unit length = m
Velocity = v
(mass of water pumped in one second)
M = mv
\(\text { Kinetic energy }=\frac{1}{2} M v^{2}\)
\(=\frac{1}{2} m v^{3}\)