JEE Physics Notes: Work, Energy and Power

Physics Notes on Work, Energy and Power

  • Work:-  Work done W is defined as the dot product of force F and displacement s. SI Unit of Work is Joule.

Here θ is the angle between  and  .

  • 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.

  • 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 mv2

Here m is the mass of the body and is the speed of the body.

  • Potential Energy (U):- Potential energy of a body is defined as, U = mgh

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.

U = mgh

  • Relation between Kinetic Energy (K) and momentum  (p):-

K = p2/2m

  • 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 .

W = ½ mv2 – ½ mu2

= 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, K+UIn accordance to law of conservation of energy, the total mechanical energy of the system always remains constant.

So, mgh + ½ mv2 = constant

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

So, E = U+K = constant

Or, Ui+Ki = Uf+Kf

Or, dU = -dK

Speed of particle v in a central force field:

v = √2/m [EU(x)]

  • Conservation of linear momentum:-

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

So, pf = pi

  • 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

v2 – v1/u1 – u2

Case (i) For perfectly elastic collision, e = 1. Thus, v2 – v1 = u1 – u2. This signifies the relative velocities of two bodies before and after collision are same.

Case (ii) For inelastic collision, e<1. Thus, v2 – v1 < u1 – u2. 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, v2 – v1 =0, or v2 = v1. 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:-

Elastic collision in One Dimension

After collision, the velocity of two body will be,

v1 = (m1m2/ m1+m2)u1 + (2m2/ m1+m2)u2


v2 = (m2m1/ m1+m2)u2 + (2m1/ m1+m2)u1


When both the colliding bodies are of the same mass, i.e., m1 = m2, then,

v1 = u2 and v2 = u1


When the body B of mass m2 is initially at rest, i.e., u2 = 0, then,

v1 = (m1m2/ m1+m2)u1 and v2 = (2m1/ m1+m2)u1

(a) When  m2<<m1, then, v1 = u1 and v2 = 2u1

(b) When  m2=m1, then, v1 =0  and v2 = u1

(c) When  m2>>m1, then, v1 = -u1 and v2 will be very small.

  • Inelastic collision:- In an inelastic collision, only the quantity momentum is conserved but not kinetic energy.

v = (m1u1+m2u2) /(m1+m2)


loss in kinetic energy, E = ½ m1u12+ ½ m2u22 – ½ (m1+ m2)v2


E= ½ (m1u12 + m2u22) – ½ [(m1u1+ m2u2)/( m1+ m2)]2

m1 m(u1u2)2 / 2( m1 m2)

  • Points to be Noted:-

(i) The maximum transfer energy occurs if m1m2

(ii) If Ki is the initial kinetic energy and Kf is the final kinetic energy of mass m1, the fractional decrease in kinetic energy is given by,

Ki – KKi = 1- v12/u21

Further, if m2 = nm1 and u2 = 0, then,

Ki – KKi = 4n/(1+n)2

  • Conservation Equation:

(i)  Momentum – m1u1+m2u2 = m1v1+m2v2

(ii) Energy – ½ m1u12+ ½ m2u22 = ½ m1v12+ ½ m2v22

  • 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.

F = – (dV/dr)

  • The line integral of a conservative force around a closed path is always zero.

\oint \vec{F}.d\vec{r} = 0

  • Spring potential energy (Es):- It is defined as,

Es = ½ kx2

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,

d2U/dx2 = +ve

(c) For unstable equilibrium,

U(x) = maximum

dU/dx = 0

d2U/dx2 = -ve

(d) For neutral equilibrium,

U(x) = constant

dU/dx = 0

d2U/dx2 = 0

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