Samacheer Class 11 Physics - Work, Energy and Power

The term work is used in diverse contexts in daily life. It refers to both physical as well as mental work. In fact, any activity can generally be called work. But in Physics, the term work is treated as a physical quantity with a precise definition. Work is said to be done by the force when the force applied to a body displaces it.

The energy possessed by the body by virtue of its position is called potential energy.
The various types of potential energies are
1) Gravitational potential energy:
The energy possessed by the body due gravitational force gives gravitational potential energy
u = mgh.
where
u → Gravitational potential energy
m → Mass of the body
g → acceleration due to gravity
h → displacement produced
2) Elastic potential energy:
The energy due to spring force and other similar forces give rise to elastic potential energy
u = 1/2² Where
U → elastic potential energy
K → spring constant
x → elongation produced
3) Electrostatic potential energy
The energy due to electro static force on charges give rise to electrostatic potential energy
U = K \(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}\) where
K = \(\frac{1}{4 \pi \varepsilon_{0}}\) Constant
q 1, q 2 – magnitude of charges
d – displacement made by any one of the charges or by both charges.

Conservative forces:
A Force is said to be a conservative force if work done by or against the force in moving the body depends only on the initial and final positions of the body and net depend on the nature of the path followed between the two positions
Examples:
Elastic spring force, electrostatic force, magnetic force, gravitational force.
Non-conservative force:
A force is said to be non-conservative if the work done by or against the force in moving a body depends on the path between initial and final positions. This means the value of work-done is different in different paths.
Examples:
* Frictional forces are non-conservative forces as the work done against friction depends on the length of the path moved by the body.
* The force due to air resistance, viscous forces are also non conservative forces because work done by or against these forces depends upon the velocity of motion.

In a collision, the total K.E of the bodies before collision is equal to the total K.E. of the bodies after collision, then it is an elastic collision.
Total K.E. before collision = Total K.E. after collision
Inelastic collision:
In a collision, the total K.E. of the bodies before collision is not equal to total K.E. after collision then it is called as inelastic collision Even though K.E. is not conserved but total energy is conserved. After collision of the two colliding bodies stick together such collision are called as perfectly inelastic a plastic collision.

a) Coefficient of restitution
b) Power
c) Law of conservation of energy
d) Loss of Kinetic energy in inelastic collision
a) Coefficient of restitution:
Coefficient of restitution defined as the ratio of velocity of separation (after collision) to velocity of approach (before collision)
The coefficient of restitution =
b) Power:
Power is defined as the rate of work done or energy delivered
P = \(\frac { W }{ t }\) = \(\vec{F}\).\(\vec{V}\)
c) Law of conservation of energy:
Law states that “Energy can neither be created nor destroyed. It may be transformed from one to another but the total energy of an isolated system remains constant”
d) Loss of kinetic energy in inelastic collision
The difference in total K.E. before collision and total K.E. after collision’s is equal to loss of K.E. during collision.
∆Q = Total K.E. before collision – Total K.E. after collision.
III. Long Answer Questions:

a) Work done by a constant force:
When a constant force ‘F’ acts on a body the small work done (dw) by the force in producing a small displacement ‘dr’ is given by dw = (F cos θ) dr.
The total W.D in producing a displacement from initial position r i, to final position r f is,
The graphical representation of the W.D by constant force is shown below
The area under the graph shows the work done by constant force.
b) Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dw) by the force in producing a small displacement dr is given by dw = (F cos θ) dr [ Here F cos 0 is the component of variable forces]. where F and 0 one variables To total W.D for the displacement from initial position r i to final position r f is given by the relation.
A graphical representation of the work done by a variable force is shown below. The area under the graph gives the W.D. by variable force.

Law:
Work done by a force on the body changes the kinetic energy of the body, ie change in K.E. = work done. This is called work energy theorem.
Proof:
Consider a body of mass m at rest on a frictionless horizontal surface. The work done (W) done by the constant force (F) for displacement (S) in the same direction is W = FS → (1)
The constant force is given by F = ma → (2)
We know that v² = u² + 2a
W = 1/2 mv² – 1/2 mu²
Here the term 1/2 mv² indicates K.E.
1/2 mv² – 1/2mu² = ∆K (change in K.E.)
∵ W = ∆K
Hence proved
Examples:
* A moving hammer drives a nail into the wood. Being in motion, it has K.E. or ability to do work.
* A fast moving stone can break a windowpane. The stone has K.E. due to its motion and so it can do work.
* The kinetic energy of air is used to run windmills.

Power is defined as the rate of work done or energy delivered. P = \(\frac { w }{ t }\)
Relation between power and velocity:
The work done by a force F for a displacement \(\overline{dr}\) is dw = \(\overline{F}\).\(\overline{dr}\)
Examples:
1) A 100 Watt bulb consumes 100 joule of electrical energy in one second
2) Electrical motor supply enough power to bring water from a bore well.

Consider two elastic bodies of masses m 1 and m 2 moving in a straight line (along positive direction) on a frictionless horizontal surface. In order to have collision assume m 1 moves faster than m 2.
Let U 1 and 1 be the initial velocities of m 1 and m 2 respectively. (u 1 > u 2 ). After collision let the masses m 1 and m 2 moves with velocities v 1 and v 2 respectively.
Incase of elastic collision both linear momentum and kinetic energies are conserved
∴ from law of conservation of linear momentum
Total momentum before (Pi)
collision = Total momentum afer collision (Pf)
m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 → (1)
m 1 (u 1 – v 1 ) = m 2 (v 2 – u 2 → (2)
Further
Total K.E. before collision (KE i ) = Total K.E. after collision (KE f )
rearranging
u 1 – u 2 = – (v 1 – v 2 ) → (5)
From this it is dear that for any elastic collision, relative speed of two elastic bodies after the collision has the same magnitude as before collision but in opposite direction.
Rewriting the above equation for v 1 & v 1
v 1 = v 2 + u 2 – u 1 → (6)
(or)
v 2 = v 1 + u 1 – u 2 → (7)
To find velocities of v 1 & v 2
Substituting (7) in (2)
m 1 (u 1 – v 1 ) = m 2 (v 1 + u 1 – u 2 – u 2 )
m 1 (u 1 – v 1 ) = m 2 (v 1 + u 1 – 2u 2 )
m 1 u 1 – m 1 v 1 = m 2 v 1 + m 2 u 1 – 2m 2 u 2
m 1 u 1 – m 2 u 1 + 2m 2 u 2 = m 1 v 1 + m 2 v 1
u 1 (m 1 – m 2 ) + 2m 2 u 2 = v 1 (m 1 + m 2 )
Case 1:
When bodies have same mass is m 1 = m 2 = m
The velocities get interchanged.
Case 2:
When both bodies have same mass m 1 = m 2 = m, but second body at rest in u 2 = 0
v 1 = 0, v 2 = u 1
After collision the first body comes to rest and the second body moves with the velocity of first body.
Case 3:
The first body very much lighter than the second body in m 1 << m 2, \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\) << 1.
the ratio \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\) = 0 and also second body at rest, (u 2 = 0)
Dividing numerator and denominator of equation 8 by m 2
Similarly dividing numerator and denominator of equation 9 by m 1
From this, the conclusion arrived is the first body which is lighter returns back (rebounds) in opposite direction with the same initial velocity as it has a negative sign. The second body since it has heavier mars continues to remain at rest even after the collision.
Case 4:
The second body is very much lighter than the first body.
m 2 << m 1 then the ratio = 0 \(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}\) and also if the target is at rest ie second body at rest (u 2 = 0)
Dividing equation (8) both the numerator and denominator by m 1
This shows that the first body which is heavier continues to move with same velocity and the second body which is lighter will move with twice the initial velocity of the first body, ie lighter body is thrown away from the point of collision.

In a collision, the total K.E. on the bodies before collision is not equal to the total K.E. after collision then it is called as inelastic collision, i.e
Total K.E after collision ≠ Total K.E. before collision
Whereas in case of elastic collision Total K.E. before collison is equal to total K.E after collision.
Example: Collision between two vehicles, collision between a ball and floor.
IV. Numerical Problems:

F = 30N
m = 2kg
s = 10 m
g = 10 ms -2
θ = 0
W.D = ?
W.D = \(\overline{F}\).\(\overline{S}\) = FS cos θ
W.D = 30 x 10
= 300 J

U 1 = 5 ms -1
θ = 60°
e = 0.5
v = ?
Initial momentum = final momentum along the original line of m of con.
∵ Coefficient of restitution is 0.5 (less than 1) the collision is inelastic
Applying component of velocities. The x component of velocity is
u sin θ = v sin Φ → (1)
But the magnitude of y component is not same using coefficient of restitution

mass of bullet m 1 = 20 x 10 -3 kg
mass of pendulum m 2 = 5kg
Let the speed of the bullet = u 1
∵ The pendulum at rest u 2 = 0
h = 10 x 10 -2 m
Let v be the common velocity after the bullet embedded inside the bob.
This common velocity ‘v’ is the initial velocity of combined bullet and bob.
W.K.T
v² – u² = 2as
Here v = 6; u = 0, a = – g s = h
V. Conceptual Questions:

In inelastic collision total energy is only conserved but kinetic energy is not conserved. A part of kinetic energy is converted into some other form of energy such as sound, heat energy.
Note: The linear momentum is also conserved.

Coulomb force is acting in between the charged particles Internal force is a conservative force. If no external forces act or the work done by external forces is zero then the mechanical energy of the system and also total linear momentum also remains constant.
11th Physics Guide Work, Energy and Power Additional Important Questions and Answers
I. Multiple choice questions: