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**Work**

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

$dW$=$F.dS$(Scalar quantity)

also,$W=FScosθ$ where $θ$is the angle between $F$and $S$.

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.)=$21 $$mv_{2}$

also, K.E. =$2mP_{2} $ where P is the momentum.

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

Gravitational P.E. = $R+h−GMm $=$R+h−gR_{2}m $ as $(g=R_{2}GM )$

If $U_{0}$is the potential energy on the ground then,

For small change in height: P.E. changes as = $U−U_{0}$

= $(R+h−gR_{2}m )$-$(R−gR_{2}m )$

=$mgh(R+hR )$

For h<<R, $ΔU$=mgh

For h>>R, $ΔU$=mgR

Potential Energy for stretched spring is the work done=$21 kX_{2}$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.

$F.S=21 m(V_{f}−V_{i})$................(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=dS−dU $

$U=−∫F.dS$

**Power: **Its the rate of doing work.

power=$time takenwork done $=$tF.S $=$F.V$=$FVcosθ$ is the scalar quantity.

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

1 Hp=746 Watt

Average power P =$∫_{0}dt∫_{0}P.dt $

**Some important solutions**

*If a uniform chain of length $l_{0}=nd $ hanging, then work done required to put chain in table is given

$W=2n_{2}Mgl =2lMgl_{0} $

solution

$W=[(lM ×l_{0})×g]×2l_{0} $

W= force * displacement

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

$=mgh(1−dρ )$

*When two bodies strikes and moves with common velocity $V=m_{1}+m_{2}m_{1}u_{1}+m_{2}u_{2} $ and loss in their kinetic energy is $21 m_{1}+m_{2}m_{1}m_{2} (u_{1}−u_{2})_{2}$

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

solution

$cosθ=LL−h $

h can be find by

$mv_{1}=(m+M)v$ where v is the common velocity.

so, $v=m+Mmv_{1} $

then

$0=v_{2}−2gh$

$∴h=2g1 (m+Mmv_{1} )_{2}$