Physics : Work, Energy and Power : Book Back Important Questions, Answers, Solutions : Short Questions and Answer

__Work, Energy and Power (Physics)__

__Short Answer Questions__

**1. Explain how the definition of work in physics is different from general percep-tion.**

●
Work refers to both physical as well as mental work. In fact, any activity can
generally be called as work.

●
But in physics, the term work is treated as physical quantity with precise
definition.

●
Work is said to be done by the force when the force applied on the body
displaces it.

**2. Write the various types of potential energy. Explain the formulae.**

i)
The energy possesed by the body due to gravitational force gives rise to
gravitational potential energy

u
= mgh

Where,
m is mass of the body

g
is acceleration due to gravity

h
is the height of the body above the ground.

ii)
The energy due to spring force and other similar forces gives rise to elastic
potential energy.

U
= 1/2 k(x_{f}^{2} – x_{i}^{2})

Where,
k is the force constant

x_{i}
is the initial position of the spring

x_{f}
is the final position of the spring

iii)
The energy due to electro static force on charge gives rise to electrostatic
potential energy.

U
= q_{1}q_{2} / 4πε_{o}r

Where,
q_{1}, q_{2} are the point charges

r
is the distance between two point charges

ε_{o
}is the permitivity of free space

ε_{o
}= 8.854 × 10^{-12} C^{2}N^{-1}m^{-2}

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**3. Write the differences between conservative and Non-conservative forces. Give two examples each.**

**Conservaive
forces **

1.
Work done is independent of the path

2.
Work done in a round trip is zero

3.
Total energy remains constant

4.
Work done is completely recoverable

5.
Force is the negative gradient of potential

6.
Examples:

i.
Elastic spring force

ii.
Electrostatic force

iii.
Magnetic force

iv.
Gravitational force

**Non
- Conservaive forces**

1.
Work done depends upon the path

2.
Work done in a round trip is not zero

3.
Energy is dissipated as heat energy

4.
Work done is not completely recoverable

5.
No such relation exists.

6.
Examples:

i.
Force due to air resistance

ii.
Viscous force.

**4. Explain the characteristics of elastic and inelastic collision.**

**Elastic
collision **

1.
Total momentum is conserved

2.
Total kinetic energy is conserved

3.
Forces involved are conservative forces

4.
Mechanical energy is not dissipated

**In
elastic collision**

1.**
**Total momentum is conserved

2.
Total kinetic energy is not conserved

3.
Forces involved are non-conservative forces

4.
Mechanical energy is dissipated into heat, light sound etc.

**5. Define the following**

**a) Coefficient of restitution**

**b) Power**

**c) Law of conservation of energy**

**d) loss of kinetic energy in inelastic collision.**

**a)
Co-efficient of restitution**

It
is defined as the ratio of velocity of separation (relative veloctiy) after
collision to the velocity of approach (relative velocity) before collision.

e = Velocity of separation (after collision) /
Velocity of approach (before collision);

e = (v_{2}−v_{1}) / (u_{1}−u_{2})

**b)
Power**

It
is defined as the rate of wrok done (or) energy delivered.

Power
(P) = Work done (w) / time taken (t)

P
= w / t

**c)
Law of conservation of energy**

It
states that energy can neither be created nor be destroyed. It may be transformed
from one form to another but total energy of an isolated system remains
constant.

**d)
Loss of kinetic energy in inelastic collision**

●
In perfect inelastic collision, the loss in kinetic energy during collision is
transformed to another form of energy like sound, thermal, heat, light etc.

●
Let KE_{i} be the total kinetic enrergy before collision and KE_{f}
be the total kinetic energy after collision.

●
Total kinetic energy before collision

KE_{1
}= 1/2 m_{1}u_{1}^{2} + 1/2 m_{2}u_{2}^{2}
………….(1)

●
Total kinetic energy after collision

KE_{f
}= 1/2 (m_{1}+m_{2})v^{2} …………………(2)

●
Then the loss of kinetic energy, ∆Q = KE_{i} − KE_{f}

∆Q
= 1/2 m_{1}u_{1}^{2} + 1/2 m_{2}u_{2}^{2}
− 1/2 (m_{1}+m_{2})v^{2 }………….(3)

●
Substituting equation *v* = [ m_{1}u_{1} + m_{2}u_{2}
] / (m_{1}+m_{2}) in equafion (3) and On simpiytying, we get loss of K.E ∆Q =
1/2 (m_{1}m_{2} / [m_{1}+m_{2}] ) (u_{1}–u_{2})^{2}

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