Work Energy Theorem And Its Applications

WORK KINETIC ENERGY THEOREM

It is possible to relate the work done by all the forces on a body (or a system) to the change in kinetic energy of the body (or, the system).

Consider a rigid body of acted upon by forces ${\vec{F}}_1,$${\vec{F}}_2$${\vec{F}}_3$, . . . moving with a velocity $\vec{v}$ which is in general a function of time. Newton's second law gives us :

m$\frac{d\vec{v}}{dt}={\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3+.......={\vec{F}}_{net}$ . . . (1)

where RHS represents the resultant (net) of all the forces ${\vec{F}}_1,{\vec{F}}_2$, . . . . .etc.

Taking the dot product of both sides of equation (1) by $\vec{v}$ dt (=d$\vec{r}$, since $\vec{v}=\frac{d\vec{r}}{dt}$), we get,

m$\vec{v}$. $\frac{d\vec{v}}{dt}$dt = ${\vec{F}}_1.d\vec{r}+{\vec{F}}_2.dr$ + . . . . = ${\bar{F}}_{net}$. d$\vec{r}$

or, m$\bar{v}.d\vec{v}$ = ${\vec{F}}_1.dr+{\vec{F}}_2.d\vec{r}$ + . . . . (2)

Now, $\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$, and $d\vec{v}$ = dv_x$\hat{i}$ + dv_y$\hat{j}$ + dv_z$\hat{k}$; so,

m (v_x dv_x + v_ydv_y + v_zdv_z) = ${\vec{F}}_{net}$.d$\vec{r}$

or, m$\underset{A}{\overset{B}{\int }}{v}_{x}d{v}_x+\underset{A}{\overset{B}{\int }}{v}_{y}d{v}_y+\underset{A}{\overset{B}{\int }}{v}_{z}d{v}_z$ = $\underset{A}{\overset{B}{\int }}{\vec{F}}_{net}.d\vec{r}$, if the motion takes place from A to B, we get by integration

or, m${\left(\frac{v_x^2+v_y^2+v_z^2}{2}\right)|}_A^B=\underset{A}{\overset{B}{\int }}{\vec{F}}_{net}.d\vec{r}$

or, $\frac{1}{2}m(v_B^2-v_A^2)=\underset{A}{\overset{B}{\int }}{F}_{net}.d\vec{r}$

or, $\frac{1}{2}$m$v_B^2-\frac{1}{2}m{v}_A^2=\underset{A}{\overset{B}{\int }}{\vec{F}}_{net}.d\vec{r}$ . . . (3)

W_net = $\Delta$KE

Conservation of Energy

An interesting case of the work-energy theorem occurs when all the forces acting on a body are conservative. In this case, one can define a potential energy for each of these forces :

U₁ = -$\underset{P_1,ref}{\int }{\vec{F}}_1.d\vec{r}$, U₂ = -$\underset{P_2,ref}{\int }{\vec{F}}_2.d\vec{r}$, . . etc.

Where P_{1, ref}, P₂,_ref are the reference points for each force.

The work -energy theorem can now be re-written, by using the relations,

dU₁ = -${\vec{F}}_1.d\vec{r}$, dU₂ = -${\vec{F}}_2.d\vec{r}$, . . .

m$\vec{v}$. $d\vec{v}$ = - dU₁ - dU₂ . . .

or, m$\vec{v}.d\vec{v}$ + dU₁ + dU₂ + . . .. = 0

Integrating, as before,

$\frac{1}{2}{m{v}^2|}_A^B$ + ${(U_1+U_2+......|}_A^B$ = 0

or, $\left(\frac{1}{2}m{v}_B^2-\frac{1}{2}m{v}_A^2\right)+(U_1^B-U_1^A+U_2^B-U_2^A+.....)=0$

or, $\frac{1}{2}m{v}_B^2+U_1^B+U_2^B+.....=\frac{1}{2}m{v}_A^2+U_1^A+U_2^A+....$

This result states that the total energy,

E_total = E_kinetic + E_potential = $\frac{1}{2}$mv² + U₁ + U₂ + . . . . . is conserved, when all the forces are conservative.

MOTION IN A VERTICAL CIRCLE

A particle of mass m is attached to a light and inextensible string. The other end of the string is fixed at O and the particle moves in a vertical circle of radius r equal to the length of the string as shown in the figure.

Consider the particle when it is at the point P and the string makes an angle q with the downward vertical. Forces acting on the particle are

T = tension in the string along its length

mg = weight of the particle vertically downward.

Hence net radial force on the particle is

$F_R=T-mgcos\theta$ . . . (1)

Since F_R = $\frac{m{v}^2}{R}$, where v = speed of the particle at P; R = radius of the circle

Here R = l (length of the string)

$F_R=T-mgcos\theta$

$\Rightarrow T=\frac{m{v}^2}{R}mg\cos \theta$ . . . (2)

Since speed of the particle decreases with height. Hence tension is maximum at the bottom, where $\cos \theta =1\left(as\theta =0\right)$

$\Rightarrow$ T_max = $\frac{m{v}^2}{l}+mg$ . . . (3)

T_min = $\frac{m{v}^{\prime }{}^2}{l}-mg$ at the top (4)

Here $v^{\prime }$ = speed of the particle at the top.

CRITICAL VELOCITY

At the top, tension is given by

T = $\frac{m{v}_T^2}{\ell }-mg$

where v_T = speed of the particle at the top. (tangential)

$\Rightarrow$ $\frac{m{v}_T^2}{\ell }=T+mg$

For v_T to be minimum, T $\approx$ 0

$\Rightarrow$ v_T = $\sqrt{g\ell }$ . . . (5)

If v_B be the critical velocity of the particle at the bottom, then from conservation of energy

mg(2$\ell$) + $\frac{1}{2}m{v}_T^2=0+\frac{1}{2}m{v}_B^2$

Q v_T = $\sqrt{g\ell }$

$\Rightarrow$ 2mg$\ell$ + $\frac{1}{2}$mg$\ell$ = $\frac{1}{2}m{v}_B^2$

$\Rightarrow$ v_B = $\sqrt{5g\ell }$ (6)

Illustration 1.

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed $\sqrt{g\ell }$. Find the speed of the particle and the inclination of the string to the vertical at that instant when the tension in the string is equal to the weight of the particle.

Solution :

$F_R=T-mgcos\theta$ . . . (i)

If v be the speed of the particle at B, then

F_R = $\frac{m{v}^2}{\ell }$ . . . (ii)

From (i) and (ii), we get

$T-mgcos\theta$= $\frac{m{v}^2}{\ell }$ . . . (iii)

Since at B, T = mg

$\Rightarrow$ $mg(1-cos\theta )$= $\frac{m{v}^2}{\ell }$

$\Rightarrow$ v² = g$\ell$ $(1-cos\theta )$ . . . (iv)

Conserving the energy of the particle at point A and B, we have

$\frac{1}{2}m{v_o}^2=mg\ell \left(1-\cos \theta \right)+\frac{1}{2}m{v}^2$

Where v_o = $\sqrt{g\ell }$ and v = $\sqrt{g\ell \left(1-\cos \theta \right)}$

$\Rightarrow$ g$\ell$= 2g$\ell$$(1-cos\theta )$ + g$\ell$ $(1-cos\theta )$

$\Rightarrow$ $cos\theta =2/3$ . . . (v)

Putting the value of $\cos \theta$ in equation (iv) we get

v = $\sqrt{\frac{g\ell }{3}}$