The lighter body has more K.E. as K.E. = \(\frac { { p }^{ 2 } }{ 2m } \) and for constant p, K.E.\(\propto \frac { 1 }{ m } \) .

Work done, \(W-\int^{x_f}_{x_i}F(x)dx=k\int_0^4x^2 dx={64\over 3}Nm\)

1 electrical unit = 1 kWh = 1\(\times\)(103 W)\(\times\)(3600 s)
1 electrical unit = 3600\(\times\)103 W s
1 electrical unit = 3.6\(\times\)10⁶ J
1 kWh = 3.6\(\times\)10⁶ J

Linear momentum.

A force is said to be non-conservative if the work done by or against the force in moving a body depends upon the path between the initial and final positions. This means that the value of work done is different in different paths.

Tension in the string is always 90° to the displacement. So, work done by the tension is zero.

Here, s=0
ஃ W = Fs cosθ = 0

The energy is dissipated in the form of heat. The heat energy so produced is not available for work.

As P.E. = mgh \(\Rightarrow\) P.E., \(\infty\) h. So the graph of P.E. verses height is a straight line as shown.

(i) If the object does not move at constant velocity, then it will have different velocities at the initial and final locations.
(ii) According to work-kinetic energy theorem, the external force will impart some extra kinetic energy.
(iii) But we associate potential energy to the forces like gravitational force, spring force and coulomb force.
(iv) So the external agency should not impart any kinetic energy when the object is taken from initial to final location.

(i) It is possible when there is another force which acts exactly opposite to the external applied force.
(ii) They both cancel each other and the resulting net force becomes zero, hence the object moves with zero acceleration.

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

Final values of potential energy, kinetic energy and total energy are measured at the height h.
By law of conservation of energy, the initial and final total energies are the same.
\(\frac { 1 }{ 2 } { mv }_{ 0 }^{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }+mgh\)
\({ v }_{ 0 }^{ 2 }={ v }^{ 2 }+2gh\)
\(v=\sqrt { { v }_{ 0 }^{ 2 }-2gh } \)

Mass of a bob = m
Length of the rod = r
From the law of conservation of energy,
\(\frac{1}{2} m v_{1}^{2} =2 m g r+\frac{1}{2} m v_{2}^{2}  \)
\(\therefore \frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2}\right) =2 m g r  \)
\(\therefore v_{1}^{2}-v_{2}^{2} =4 g r  \)
\(\text { If } v_{2} =0 \text { then }  \)
\(v_{1}^{2} =4 g r  \)
\(\therefore v_{1} =\sqrt{4 g r} m s^{-1}\)

(a) Coefficient of restitution
It is defined as the ratio of velocity of separation (relative velocity) after collision to the velocity of approach (relative velocity) before collision.
\(e=\frac{v_2-v_1}{u_1-u_2}\)