(i) A body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius \(\vec{r}\) of the circular path.
(ii) The motion of the body by taking the free body diagram (FBD) at a position where the position vector \(\vec{r}\) makes an angle e with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, In the tangential direction,

mg sinθ = ma_t
mg sinθ = -m \((\frac{dv}{dt})\)
where, a_t = -\((\frac{dv}{dt})\) is tangential retardation
In the radial direction,
T - mg cosθ = m a_r
T - mg cosθ = \(\frac{mv^{2}}{r}\)
where, a_r = \(\frac{v^{2}}{r}\) is the centripetal acceleration.

(i) When an object is thrown upwards its kinetic energy goes on decreasing and consequently its potential energy keeps increasing (neglecting air resistance).
(ii) When it reaches the highest point its energy is completely potential. Similarly, when the object falls back from a height its kinetic energy increases whereas its potential energy decreases.
(iii) When it touches the ground its energy is completely kinetic. At the intermediate points the energy is both kinetic and potential as shown in Figure.
(iv) When the body reaches the ground the kinetic energy is completely dissipated into some other form of energy like sound, heat, light and deformation of the body etc.
(v) In this example the energy transformation takes place at every point. The sum of kinetic energy and potential energy i.e., the total mechanical energy always remains constant, implying that the total energy is conserved. This is stated as the law of conservation of energy.

(vi) The law of conservation of energy states that energy can neither be created nor destroyed. It may be transformed from one form to another but the total energy of an isolated system remains constant.
(vii)The figure Illustrates that, if an object starts from rest at height h, the total energy is purely potential energy (U=mgh) and the kinetic energy (KE) is zero at h. When the object falls at some distance y, the potential energy and the kinetic energy are not zero whereas, the total energy remains same as measured at height h. When the object is about to touch the ground, the potential
energy is zero and total energy is purely kinetic.

(i) The potential energy possessed by a spring due to a deforming force which stretches or compresses the spring is termed as elastic potential energy. The work done by the applied force against the restoring force of the spring is stored as the elastic potential energy in the spring.
(ii) Consider a spring-mass system. Let us assume a mass, m lying on a smooth horizontal table as shown in Figure. Here, x = 0 is the equilibrium position. One end of the spring is attached to a rigid wall and the other end to the mass.
               
(iii) As long as the spring remains in equilibrium position, its potential energy is zero. Now an external force \(\overrightarrow { { F }_{ a } } \) is applied so that it is stretched by a distance (x) in the direction of the force.
(iv) There is a restoring force called spring force \(\overrightarrow { { F }_{ s } } \) developed in the spring which tries. to bring the mass back to its original position. This applied force and the spring force are equal in magnitude but opposite in direction i.e \(\overrightarrow { { F }_{ a } } \) = - \(\overrightarrow { { F }_{ s } } \) According Hooke's law, the restoring force developed in the spring is
\(\overrightarrow { { F }_{ s } } \) = -k\(\overrightarrow { x } \)
(v) The negative sign in the above expression implies that the spring force is always opposite to that of displacement \(\overrightarrow { x } \) and k is the force constant. There, fore applied force is \(\overrightarrow { { F }_{ a } } \) = +k\(\overrightarrow { x } \) The positive sign implies. that the applied force is in the direction of displacement \(\overrightarrow { x } \). The spring force is an example of variable force as it depends on the displacement \(\overrightarrow { x } \). Let the spring be stretched to a small distance d\(\overrightarrow { x } \). The work done by the applied force on the spring to stretch it by a displacement \(\overrightarrow { x } \) is stored as elastic potential energy.
 \(U=\int \overrightarrow { { F }_{ a } } .d\overrightarrow { r } =\overset { x }{ \underset { 0 }{ \int } } \left| \overrightarrow { { F }_{ a } } \right| \left| d\overrightarrow { r } \right| \cos { \theta } \)
\(=\overset { x }{ \underset { 0 }{ \int } } { F }_{ a }dx\cos { \theta } \)
(vi) The applied force \(\overrightarrow { { F }_{ a } } \) and the displacement d\(\overrightarrow { { r } } \) (i.e., here dx ) are in the same direction. As, the initial position is taken as the equilibrium position or mean position, x = 0 is the lower limit of integration.
\(U=\overset { x }{ \underset { 0 }{ \int } } Kxdx\)
\(U={ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ x }\)
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }\)
(vii) If the initial position is not zero, and if the mass is changed from position x_i to x_f then the elastic potential energy is
\(U=\frac { 1 }{ 2 } k\left( { x }_{ f }^{ 2 }-{ x }_{ i }^{ 2 } \right) \)
From equations (1) and (2), we observe that the' potential energy of the stretched spring depends on the force constant k and elongation or compression x.

(i) The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from ground to that height h with constant velocity.
(ii) Consider a body of mass m being moved from ground to the height h against the gravitational force as shown in Figure.
            
iii) The gravitational force \({ \overrightarrow { F } }_{ g }\) acting on the body is, \({ \overrightarrow { F } }_{ g }=-mg\hat { j } \) (as the force is' in y-direction, unit vector is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force \({ \overrightarrow { F } }_{ a }\) equal in magnitude but opposite to that of gravitational force \({ \overrightarrow { F } }_{ a }-{ \overrightarrow { F } }_{ g }\) . This implies that \({ \overrightarrow { F } }_{ a }=-mg\hat { j } \)
(iv) The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.
(v) 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.
\(U=\int \overrightarrow { { F }_{ a } } .d\overrightarrow { r } =\int \left| \overrightarrow { { F }_{ a } } \right| \left| d\overrightarrow { r } \right| \cos { \theta } \)
(vi) Since the displacement and the applied force are in the same upward direction, the angle between them, \(\theta ={ 0 }^{ o }\)
Hence, \(\cos { { \theta }^{ o } } =1\) and \(\left| \overrightarrow { { F }_{ a } } \right| =mg\) and
\(=\left| d\overrightarrow { r } \right| dr.\)
\(U=\overset { h }{ \underset { 0 }{ \int } } dr\)
\(U=mg{ \left[ r \right] }_{ 0 }^{ h }\) = mgh.

(i) Consider an object of mass m moving with a velocity \(\vec{v}\). Then its linear momentum is \(\vec{p}=m\vec{v}\) and Its kinetic energy, KE \(\frac{1}{2}mv^{2}\)
\(KE=\frac{1}{2}mv^{2}=\frac{1}{2}m(\vec{v}.\vec{v})\)    --- (1)
(ii) Multiplying both the numerator and denominator of equation (1) by mass, m
\(KE=\frac{1}{2}\frac{m^{2}(\vec{v}.\vec{v})}{m}\)
=\(\frac{1}{2}\frac{(m\vec{v}).(m\vec{v})}{m}[\vec{p}=m\vec{v}]\)
=\(\frac{1}{2}\frac{\vec{p}.\vec{p}}{m}\) =\(\frac{\vec{p}^{2}}{2m}\)
KE = \(\frac{p^{2}}{2m}\)   ......(2)
(ii) where \(|\vec{p}|\) is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by
\(|\vec{p}|\) = p = \(\sqrt{2m(KE)}\)   .....(3)
(iv) Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars.