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**Work:-**Work done W is defined as the dot product of force F and displacement s.

Here θ is the angle betweenand.

Work done by the force is positive if the angle between force and displacement is acute (0°<θ<90°) as cos θ is positive. This signifies, when the force and displacement are in same direction, work done is positive. This work is said to be done upon the body.- When the force acts in a direction at right angle to the direction of displacement (cos90° = 0), no work is done (zero work).
- Work done by the force is negative if the angle between force and displacement is obtuse (90°<θ<180°) as cosθ is negative. This signifies, when the force and displacement are in opposite direction, work done is negative. This work is said to be done by the body.
**Work done by a variable force:-**

If applied force F is not a constant force, then work done by this force in moving the body from position A to B will be,

Here ds is the small displacement.**Units:**The unit of work done in S.I is joule (J) and in C.G.S system is erg.

1J = 1 N.m , 1 erg = 1 dyn.cm**Relation between Joule and erg:**- 1 J = 107 erg**Power:-**The rate at which work is done is called power and is defined as,

P = W/t = F.s/v = F.v

Here s is the distance and v is the speed.**Instantaneous power in terms of mechanical energy:**- P = dE/dt**Units:**The unit of power in S.I system is J/s (watt) and in C.G.S system is erg/s.**Energy:-**

1) Energy is the ability of the body to do some work. The unit of energy is same as that of work.

2) Kinetic Energy (K):- It is defined as,

K= ½ mv^{2}

Here m is the mass of the body and v is the speed of the body.**Potential Energy (U):-**Potential energy of a body is defined as, U = mgh

Here, m is the mass of the body, g is the free fall acceleration (acceleration due to gravity) and h is the height.**Gravitational Potential Energy:-**An object’s gravitational potential energy U is its mass m times the acceleration due to gravity g times its height h above a zero level.

In symbol’s,

U = mgh**Relation between Kinetic Energy (K) and momentum (p):-**

K = p^{2}/2m- If two bodies of different masses have same momentum, body with a greater mass shall have lesser kinetic energy.
- If two bodies of different mass have same kinetic energy, body with a greater mass shall have greater momentum.
- For two bodies having same mass, the body having greater momentum shall have greater kinetic energy.
**Work energy Theorem:**- It states that work done on the body or by the body is equal to the net change in its kinetic energy .**For constant force,**

W = ½ mv^{2 }– ½ mu^{2}

= Final K.E – Initial K.E- For variable force,
**Law of conservation of energy:-**It states that, “Energy can neither be created nor destroyed. It can be converted from one form to another. The sum of total energy, in this universe, is always same”.- The sum of the kinetic and potential energies of an object is called mechanical energy. So, E = K+U
- In accordance to law of conservation of energy, the total mechanical energy of the system always remains constant.

So, mgh + ½ mv^{2}= constant

In an isolated system, the total energy Etotal of the system is constant.

So, E = U+K = constant

Or, U_{i}+K_{i}= U_{f}+K_{f}

Or, ?U = -?K

Speed of particle v in a central force field:

v = √2/m [E-U(x)] **Conservation of linear momentum:-**

In an isolated system (no external force ( F_{ext}= 0)), the total momentum of the system before collision would be equal to total momentum of the system after collision.

So, p_{f }= p_{i}**Coefficient of restitution (e):-**It is defined as the ratio between magnitude of impulse during period of restitution to that during period of deformation.

e = relative velocity after collision / relative velocity before collision

= v_{2 }– v_{1}/u_{1}– u_{2}**Case (i)**For perfectly elastic collision, e = 1. Thus, v_{2}– v_{1}= u_{1 }– u_{2}. This signifies the relative velocities of two bodies before and after collision are same.**Case (ii)**For inelastic collision, e<1. Thus, v_{2}– v_{1}< u_{1}– u_{2}. This signifies, the value of e shall depend upon the extent of loss of kinetic energy during collision.**Case (iii)**For perfectly inelastic collision, e = 0. Thus, v_{2}– v_{1 }=0, or v_{2}= v_{1}. This signifies the two bodies shall move together with same velocity. Therefore, there shall be no separation between them.**Elastic collision:**- In an elastic collision, both the momentum and kinetic energy conserved.**One dimensional elastic collision:-**

After collision, the velocity of two body will be,

v_{1}= (m_{1}-m_{2}/ m_{1}+m_{2})u_{1}+ (2m_{2}/ m_{1}+m_{2})u_{2}

and

v_{2}= (m_{2}-m_{1}/ m_{1}+m_{2})u_{2 }+ (2m_{1}/ m_{1}+m_{2})u_{1}**Case:I**

When both the colliding bodies are of the same mass, i.e., m_{1}= m_{2}, then,

v_{1}= u_{2}and v_{2}= u_{1}**Case:II**

When the body B of mass m2 is initially at rest, i.e., u_{2}= 0, then,

v_{1}= (m_{1}-m_{2}/ m_{1}+m_{2})u_{1}and v_{2 }= (2m_{1}/ m_{1}+m_{2})u_{1}

(a) When m_{2}<<m_{1}, then, v_{1}= u_{1}and v_{2}= 2u_{1}

(b) When m_{2}=m_{1}, then, v_{1}=0 and v_{2}= u_{1}

(c) When m_{2}>>m_{1}, then, v_{1 }= -u_{1}and v_{2}will be very small.**Inelastic collision:-**In an inelastic collision, only the quantity momentum is conserved but not kinetic energy.

v = (m_{1}u_{1}+m_{2}u_{2}) /(m_{1}+m_{2})

and

loss in kinetic energy, E = ½ m_{1}u_{1}^{2}+ ½ m_{2}u_{2}^{2}- ½ (m_{1}+ m_{2})v_{2}

or,

E= ½ (m_{1}u_{1}^{2}+ m_{2}u_{2}^{2}) – ½ [(m_{1}u_{1}+ m_{2}u_{2})/( m_{1}+ m_{2})]^{2}

= m_{1}m_{2}(u_{1}-u_{2})^{2 }/ 2( m_{1}+ m_{2})**Points to be Notice:-**

(i) The maximum transfer energy occurs if m_{1}= m_{2}

(ii) If K_{i }is the initial kinetic energy and K_{f}is the final kinetic energy of mass m_{1}, the fractional decrease in kinetic energy is given by,

K_{i }– K_{f}/ K_{i}= 1- v_{1}^{2}/u^{2}_{1}

Further, if m_{2 }= nm_{1}and u_{2}= 0, then,

K_{i}– K_{f}/ K_{i }= 4n/(1+n)^{2}**Conservation Equation:**

(i) Momentum – m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}v_{2}

(ii) Energy – ½ m_{1}u_{1}^{2}+ ½ m_{2}u_{2}^{2}= ½ m_{1}v_{1}^{2}+ ½ m_{2}v_{2}^{2}- Conservative force (F):- Conservative force is equal to the negative gradient of potential V of the field of that force. This force is also called central force.

So, F = - (dV/dr) - The line integral of a conservative force around a closed path is always zero.

So, **Spring potential energy (E**It is defined as,_{s}):-

E_{s }= ½ kx^{2}

Here k is the spring constant and x is the elongation.**Equilibrium Conditions:**

(a) Condition for equilibrium, dU/dx = 0

(b) For stable equilibrium,

U(x) = minimum,

dU/dx = 0,

d^{2}U/dx^{2}= +ve

(c) For unstable equilibrium,

U(x) = maximum

dU/dx = 0

d^{2}U/dx^{2}= -ve

(d) For neutral equilibrium,

U(x) = constant

dU/dx = 0

d^{2}U/dx^{2 }= 0**UNITS AND DIMENSIONS OF WORK, POWER AND ENERGY**

Work and Energy are measured in the same units. Power, being the rate at which work is done, is measured in a different unit.Quantity and Units/DimensionsWork (Energy)PowerDimensionML^{2}T^{-2}ML^{2}T^{-3}AbsoluteMKSJouleWattFPSft-Poundalft-poundal/secCGSergErg/sec.GravitationalMKSkg-mKg-m/secFPSft-lbft-lb/sec.CGSgm-cmGm-cm/secPractical(Other)kwh, eV, calHP, kW, MW**Conversions between Different Systems of Units**

1 Joule = 1 Newton ´ 1 m = 10^{5}dyne ´ 10^{2}cm = 10^{7}erg

1 watt = 1 Joule/ sec = 10^{7}erg/sec.

1 kwh = 10^{3}watt ´ 1 hr = 10^{3}watt ´ 3600 sec

= 3.6 ´ 10^{6}Joule

1HP = 746 watt.

1 MW = 10^{6}watt.

1 cal = 1 calorie = 4.2 Joule

1eV = "e" Joule = 1.6 ´ 10^{-19}Joule

(e = magnitude of charge on the electron in coulombs)

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