Work, Energy and Power

Work

It is done when body transverse a certain distance on action of force. It is given by:

$dW$=$\vec{F}.\overrightarrow{dS}$(Scalar quantity)

also,$W=FScos\theta$ where $\theta$is the angle between $\vec{F}$and $\vec{S}$.

Work done by frictional force is negative because it is retarding force.

SI unit of work is Joule and CGS is erg.

1 Joule =$10^7$erg

Conservative and Non-conservative force

Conservative force: The total mechanical energy is conserved and work done depends on initial and final position (dependent on path followed).So that work done on closed path is always zero. eg: Gravitational force, electrostatic force, magnetostatic force, etc.

Non-conservative force: The total mechanical energy is not conserved and work done depends on actual path followed(path function). So that the work done on closed path is never zero.eg: frictional and viscous forces

Energy

It is the capacity of the body to do work. It has same dimension of work and also scalar in quantity.

Mechanical energy is the energy due to position or configuration.

Kinetic energy (K.E.) and Potential Energy (P.E.) are mechanical energy.

Kinetic energy: It is due to the motion of the body.

If m be the mass and v be the velocity of the body while moving then

kinetic energy (K.E.)=$\frac{1}{2}$$m{v}^2$

also, K.E. =$\frac{P^2}{2m}$ where P is the momentum.

Potential energy: It is the energy due to configuration (position).

Gravitational P.E. = $\frac{-GMm}{R+h}$=$\frac{-g{R}^{2}m}{R+h}$ as $\left(g=\frac{GM}{R^2}\right)$

If $U_0$is the potential energy on the ground then,

For small change in height: P.E. changes as = $U-U_0$

= $\left(\frac{-g{R}^{2}m}{R+h}\right)$-$\left(\frac{-g{R}^{2}m}{R}\right)$

=$mgh\left(\frac{R}{R+h}\right)$

For h<<R, $\Delta U$=mgh

For h>>R, $\Delta U$=mgR

Potential Energy for stretched spring is the work done=$\frac{1}{2}k{X}^2$where k is the spring constant and X is the extension.

Work Energy theorem and conservation of Energy

Work and Energy are equivalent.

Work done = Change in K.E.

$F.S=\frac{1}{2}m(V_f^2-V_i^2)$................(i)

Conservation of energy states that the total energy remains constant.

Then considering the effect of only K.E. and P.E.

Total energy= K.E. + P.E.= K+U= let[constant(C)]

K+U=C (diff. on both sides)

dK+dU=0

dU=-dK=-Fds.............(from eqn(i))

$F=\frac{-dU}{dS}$

$U=-\int \vec{F}.\overrightarrow{dS}$

Power: Its the rate of doing work.

power=$\frac{\text{work done}}{\text{time taken}}$=$\frac{\vec{F}.\vec{S}}{t}$=$\vec{F}.\vec{V}$=$FVcos\theta$ is the scalar quantity.

Its SI unit is Watt and practical unit is Horse power(Hp)

1 Hp=746 Watt

Average power P =$\frac{{\int }_0^{t}P.dt}{{\int }_0^{t}dt}$

Some important solutions

*If a uniform chain of length $l_0=\frac{d}{n}$ hanging, then work done required to put chain in table is given

$W=\frac{Mgl}{2n^2}=\frac{Mg{l}_0^2}{2l}$

solution

$W=\left[\left(\frac{M}{l}\times l_0\right)\times g\right]\times \frac{l_0}{2}$

W= force * displacement

*Work done in pulling the body of mass m and density d in liquid of density $\rho$ through height h is

$=mgh\left(1-\frac{\rho }{d}\right)$

*When two bodies strikes and moves with common velocity $V=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$ and loss in their kinetic energy is $\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(u_1-u_2{)}^2$

*If bullet of mass m gets embedded in the block of mass M suspended by string of length L, angle made by string is $\theta$then

solution

$cos\theta =\frac{L-h}{L}$

h can be find by

$m{v}_1=(m+M)v$ where v is the common velocity.

so, $v=\frac{m{v}_1}{m+M}$

then

$0=v^2-2gh$

$\therefore h=\frac{1}{2g}{\left(\frac{m{v}_1}{m+M}\right)}^2$