Chapter – 10, Work, Power and Energy
Definition Of Work : When a force is applied on an object to displace it from its initial position then it is said to be work is done. It is the Dot Product of Force and Displacement.
Types Of Work :
1 ) Positive Work : When the applied force and displacement of the objects take place in the same direction or at acute angle , then it is said to be positive work is done. That is the applied force and displacement are parallel to each other and angle between them is zero degree and the value of cos0 is 1 i.e
W = FS
OR
W = FS cosθ
Examples :
- Kicking of a football
- Moving a chair
- pushing and moving a table
- A car moving forward
- throwing a stone
2 ) Negative Work : When the applied force and displacement of the objects take place in the opposite direction, then it is said to be negative work is done. That is the applied force and displacement are antiparallel to each other and angle between them is 180 degree and the value of cos180 is ( – 1 ) i.e
W = -FS
Examples :
- Two boys push each other
- In Tug of War when opponent team pulls hard.
- Trying to catch a heavy object, hands move down
3 ) Zero Work : When the applied force and displacement of the objects take place perpendicularly, then it is said to be zero work is done. That is the applied force and displacement are normal to each other and angle between them is 90 degree and the value of cos90 is 0 . when we apply a force on the object but it does not displace from its position, then there is zero work done .
W = 0
Examples :
- Pushing a wall
- Walking on the road
- Motion of planet on its orbit
Summary of types of work
Unit Of Work :
1 ) SI – Unit —- Joule
2 ) CGS – Unit ——— erg
Dimension Of Work :
[ W ] = [ F ] [ S ] = [ MLT^{-2} ] [ L ] = [ ML^{2}T^{-2} ]
Workdone By Variable Force :
In everyday life, most of the force are variable in nature.
Whenever we apply variable force on the object to displace it from its previous position, the workdone in this case may be termed as workdone by variable force.
Definition Of Power : The rate of doing work is called as Power.
Power = Work / time
P = dw / dt
where dw = small workdone
dt = time interval
It is the scalar product of Force and Velocity.
P = F . v
Unit Of Power :
1 ) SI – Unit : ( 1) Watt ( 2 ) J / Sec ( 3 ) N-m / sec
2 ) CGS unit : ( 1 ) gm – cm / sec ( 2 ) dyne – cm / sec
Dimension Of Power :
[ Power ] = [ Work ] / [ Time ] = [ ML^{2}T^{-2} ] / [ T ] = [ ML^{2}T^{-3} ]
Definition of energy :
The capacity of doing work is called energy.
Types Of Energy :
There are several types of energy which are given below:
- Mechanical Energy
- Chemical energy
- Atomic energy
- Light energy
- Sonic energy
- Wind energy
- Thermal energy
- Nuclear energy
- Electromagnetic energy
- Gravitational energy
- Ionization energy
- Electrical energy
- Heat energy
- Radiant energy
- Elastic energy
- Internal energy
These are not enough but in the syllabus of class – 11 physics, we will discuss only mechanical energy in detail.
Mechanical Energy :
The energy possessed by an object according to its structure, configuration, position and motion, is called as Mechanical energy .
There are two types of mechanical energy
- Kinetic energy
- potential energy
Kinetic energy :
The energy possessed by an object with respect to its motion, is termed as kinetic energy.
Examples :
1 ) A moving car
2 ) Moving fan
3 ) The arrow flying through air when the string is released on the bow.
There are basically several types of kinetic energy
Rotational kinetic energy, Translational kinetic energy, Vibrational kinetic energy, Translational – rotational kinetic energy and etc.
Expression for kinetic energy :
Let
u = initial velocity = 0 ( ∵ the object is initially at rest )
v = final velocity
a = acceleration
m = mass of the object
t = time
F = force applied on the object to displace it from its position
s = displacement
From Newton’s 2nd law of motion we have
F = ma —————————————— ( 1 )
From equation of motion we have
v^{2} = u^{2} + 2as
⇒ v^{2} = 2as
⇒ s = v^{2} / 2a
Now, workdone = force * displacement = ma * v^{2} / 2a = mv^{2} / 2
Potential energy :
The energy possessed by an object due to its configuration, structure and height is called as potential energy. This energy is associated with the objects at rest.
Examples :
1 ) A stretched string
2 ) A coiled spring
3 ) A snow pack
4 ) A stretched rubber band
5 ) A rock sitting at the edge of cliff
6 ) A raised weight
7 ) stored water in a tank or dam
Different types of potential energy:
1 ) Gravitational Potential Energy :
The energy required to bring a body of definite mass from infinity to a point in the gravitational field is called as Gravitational potential energy. In case of Earth, this is the amount of work done to raise a body to a particular height.
2 )Electrostatic potential energy :
The energy of a system of two or more charged bodies is called as Electrostatic potential energy.
3) Elastic potential energy :
When a mass attached to a spring is displaced from its equilibrium position, a restoring force is come into play. The work done against the restoring force is stored in the form of potential energy. This type of potential energy is termed as Elastic potential energy.
Expression Of The Gravitational Potential Energy :
Let an object of mass ‘ m ‘ is raised to a height of ‘ h ‘. The work done in raising the object is stored in the form of potential energy.
∴ Potential energy = Force by which the object is raised * height = Weight * height = mgh
where g = acceleration due to gravity.
P.E = mgh
Expression for the elastic potential energy
or
potential energy of a spring :
Let us assume that a very light and perfectly elastic spring is fixed at one end from a rigid support. Other end of the support is attached with a body of mass ‘m’. ( In the above figure, red circular part is assumed as the body. Let the body is displaced from it’s equilibrium position. Due to elastic property of spring, the restoring force come into play and the whole system is seemed to be in to and fro motion.
Let
displacement be = x
restoring force = F
According to Hook’s Law
F ∝ – x —————-( 1 )
F = – kx
where k = spring constant or force constant
Work done against the restoring force is given by
w = ∫ Fdx
where dx = small displacement
w = ∫ ( – kx ) dx
w = – k ∫ xdx
w = – k x^{2} / 2 + c
where ‘c’ is integration constant
when x = 0 then w = 0
0 = – k ( 0 )^{2} / 2 + c
c = 0
w = – kx^{2} / 2
this work done is stored in the form of elastic potential energy given by
∴ U = kx^{2} / 2
Principle of conservation of energy :
It states that
” Energy can not be created nor be destroyed but can be converted from one form to another ”
OR
” Energy of an isolated system remain constant ”
Conservative force : When the work done by a force in moving a body from one point to another is independent of the followed between the two points then the force is said to be conservative force.
Example : Gravitational force, elastic force and electrostatic force etc.
Properties of Conservative force :
- The work done is independent of path followed between the initial and final position of a body.
- The work done by this force depends only on the initial and final position of the body.
- The work done by this force in moving a particle along a closed path is zero.
- The work done by this force is always reversible.
- Mechanical energy of a body is always conserved under the action of conservative force.
Non conservative force : When the work done by a force in moving a body from one point to another is dependent of the followed between the two points then the force is said to be conservative force. It is also termed as dissipative force.
Example : Frictional force, force due to air resistance.
Properties of Non Conservative force :
- The work done is dependent of path followed between the initial and final position of a body.
- The work done by this force does not depend only on the initial and final position of the body.
- The work done by this force in moving a particle along a closed path is non zero.
- The work done by this force is always irreversible.
- Mechanical energy of a body is not conserved under the action of conservative force.
Collision :
an accident that happen when two bodies hit each other with force , is called as collision and these vodies are termed as colliding objects or colliding bodies.
Types of collision :
1 ) Head – on – collision : when two bodies moving in opposite direction crash into each other. The collision occurred is called as head – on – collision. This one of the most dangerous accident because the impact is doubled due to the travelling speed f each body. It is also termed as fave – to – face collision.
What
2 ) Elastic collision : The collision in which there is no loss of kinetic energy in the colliding system, is termed as elastic collision.
Properties of Elastic collision are discussed in the following way.
- Principle of conservation of linear momentum is obeyed.
- Total kinetic energy of the system is conserved.
- The total energy of the system is conserved.
- The mechanical energy is not converted into other form of energy like sound energy, light energy and heat energy etc.
- Forces involved during the interaction are conservative nature.
3 ) Inelastic collision: The collision in which the kinetic energy of the system is not conserved due to action of internal friction, is called as inelastic collision. This type of collision obeys the principle of conservation of linear momentum.
4 ) Perfectly inelastic collision : The collision in which the colliding bodies are stuck together after collision and then move with a common velocity , is called as perfectly inelastic collision.
Properties of Perfectly Inelastic collision are discussed in the following way.
- Total energy is conserved
- The linear momentum of the system is conserved.
- Kinetic energy is not conserved.
- A part of mechanical energy may be converted into other form of energy like sound energy, heat energy and light energy.
- Forces involved during the interaction are non – conservative nature.
Collision in one dimension
a ) Inelastic collision in one dimension
1 ) Expression for final velocity
Let
m_{1} = mass of first colliding object
m_{2} = mass of second colliding object
u_{1} = initial velocity of first colliding object
u_{2} = initial velocity of second colliding object
v = final velocity of the combined mass
applying principle of conservation of linear momentum
m_{1}u_{1} + m_{2}u_{2} = ( m_{1} + m_{2} )v
v = ( m_{1}u_{1} + m_{2}u_{2} ) / ( m_{1} + m_{2} )
2 ) Expression for loss in kinetic energy
Δk = m_{1}u^{2}_{1} / 2 + m_{2}u^{2}_{2} / 2 – ( m_{1} + m_{2} )v^{2} / 2
Δk = [ m^{2}_{1}u^{2}_{1} + m_{1}m_{2}u^{2}_{2} + m_{1}m_{2}u^{2}_{2} + m^{2}_{2}u^{2}_{2} – m^{2}_{1}u^{2}_{1} – m^{2}_{2}u^{2}_{2} -2m_{1}mu_{1}u_{2} ] / 2 ( m_{1} + m_{2} )
Δk = [ m_{1}m_{2}u^{2}_{1} + m_{1}m_{2}u^{2}_{2} – 2m_{1}m_{2}u_{1}u_{2} ] / 2( m_{1} + m_{2} )
Δk = [ m_{1}m_{2} ( u_{1}^{2} + u_{2}^{2} – 2u_{1}u_{2} ) ] / 2 ( m_{1} + m_{2} )
Δk = [ m_{1}m_{2} ( u_{1} – u_{2} )^{2} ] / 2 ( m_{1} + m_{2} )
# Special cases :
1 ) when u_{2} = 0 i.e the second colliding body is at rest
v = ( m_{1}u_{1} + m_{2}u_{2} ) / ( m_{1} + m_{2} )
v = m_{1}u_{1} / ( m_{1} + m_{2} )
and
Δk = [ m_{1}m_{2} ( u_{1} – u_{2} )^{2} ] / 2 ( m_{1} + m_{2} )
Δk = m_{1}m_{2}u_{1}^{2} / 2 ( m_{1} + m_{2} )
2 ) when m_{1} = m_{2} = m ( say ) i.e both the colliding bodies have equal masses
v = ( m_{1}u_{1} + m_{2}u_{2} ) / ( m_{1} + m_{2} )
v = m ( u_{1} + u_{2} ) / 2m
v = ( u_{1} + u_{2} ) / 2
and
Δk = [ m_{1}m_{2} ( u_{1} – u_{2} )^{2} ] / 2 ( m_{1} + m_{2} )
Δk = m^{2} ( u_{1} – u_{2} )^{2} / 4m
Δk = m( u_{1} – u_{2} )^{2} / 4
3 ) when m_{1} = m_{2} = m ( say ) and u_{2} = 0
v = ( m_{1}u_{1} + m_{2}u_{2} ) / ( m_{1} + m_{2} )
v = mu_{1} / 2m
v = u_{1} / 2
Δk = [ m_{1}m_{2} ( u_{1} – u_{2} )^{2} ] / 2 ( m_{1} + m_{2} )
Δk = m^{2}u_{1}^{2} / 4m
Δk = mu^{2} / 4
b ) Elastic collision in one dimension
1 ) Coefficient of restitution ( e )
Let
m_{1} = mass of first colliding object
m_{2} = mass of second colliding object
u_{1} = initial velocity of first colliding object
u_{2} = initial velocity of second colliding object
v_{1} = final velocity of the first colliding object
v = final velocity of the second colliding object
applying principle of conservation of linear momentum
m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}
m_{1}u_{1} – m_{1}v_{1} = m_{2}v_{2} – m_{2}u_{2}
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} ) ———————————- ( 1 )
applying principle of conservation of kinetic energy
m_{1}u^{2}_{1} / 2 + m_{2}u^{2}_{2} / 2 = m_{1}v^{2}_{1} / 2 + m_{2}v^{2}_{2} / 2
m_{1}u^{2}_{1} – m_{1}v^{2}_{1} = m_{2}v^{2}_{2} – m_{2}u^{2}_{2}
m_{1} ( u^{2}_{1} – v^{2}_{1} ) = m_{2} ( v^{2}_{2} – u^{2}_{2} )
m_{1} ( u_{1} – v_{1} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
by using equation ( 1 ) we get
m_{2} ( v_{2} – u_{2} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
( u_{1} + v_{1} ) = ( v_{2} + u_{2} ) ——————————— ( 2 )
( v_{2} – v_{1} ) = ( u_{1} – u_{2} )
( v_{2} – v_{1} ) / ( u_{1} – u_{2} ) = 1
e = 1
where e = ( v_{2} – v_{1} ) / ( u_{1} – u_{2} ) = coefficient of restitution
2 ) Expression for final velocity of first colliding body
from equation ( 2 ) we get
( u_{1} + v_{1} ) = ( v_{2} + u_{2} )
v_{2} = u_{1} + v_{1} – u_{2}
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} )
m_{1} ( u_{1} – v_{1} ) = m_{2} ( u_{1} + v_{1} – u_{2} – u_{2} )
( m_{1} u_{1} – m_{1}v_{1} )= ( m_{2}u_{1} + m_{2}v_{1} – 2m_{2}u_{2} )
( m_{2}v_{1} + m_{1}v_{1} ) = m_{1} u_{1} – m_{2}u_{1} + 2m_{2}u_{2}
v_{1}( m_{1} + m_{2} ) = u_{1}( m_{1} – m_{2} ) + 2m_{2}u_{2}
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} ) + 2m_{2}u_{2} / ( m_{1} + m_{2} )
3 ) Expression for final velocity of second colliding body
from equation ( 2 ) we get
( u_{1} + v_{1} ) = ( v_{2} + u_{2} )
v_{1} = ( v_{2} + u_{2} – u_{1} ) ——————————— ( 3 )
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} )
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{2} – u_{2} + u_{1} ) = m_{2} ( v_{2} – u_{2} )
( m_{1}u_{1} – m_{1}v_{2} – m_{1}u_{2} + m_{1}u_{1} ) = ( m_{2}v_{2} – m_{2}u_{2} )
( m_{2}v_{2} + m_{1}v_{2} ) = ( m_{1}u_{1}– m_{1}u_{2} + m_{1}u_{1} + m_{2}u_{2} )
v_{2}( m_{2} + m_{1}) = ( 2m_{1}u_{1 }– m_{1}u_{2} + m_{2}u_{2} )
v_{2}= 2m_{1}u_{1 } / ( m_{2} + m_{1}) _{ }– u_{2}( m_{1} – m_{2} ) / ( m_{2} + m_{1})
# special cases :
1 ) when u_{2} = 0 i.e the second colliding body is at rest
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} ) + 2m_{2}u_{2} / ( m_{1} + m_{2} )
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} )
and
v_{2 }= 2m_{1}u_{1 } / ( m_{2} + m_{1}) _{ }– u_{2}( m_{1} – m_{2} ) / ( m_{2} + m_{1})
v_{2 }= 2m_{1}u_{1 } / ( m_{2} + m_{1})
2 ) when m_{1} = m_{2} = m ( say ) i.e both the colliding bodies have equal masses
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} ) + 2m_{2}u_{2} / ( m_{1} + m_{2} )
v_{1 }= 0 + 2mu_{2} / ( 2m )
v_{1 }= u_{2}
and
v_{2 }= 2m_{1}u_{1 } / ( m_{2} + m_{1}) _{ }– u_{2}( m_{1} – m_{2} ) / ( m_{2} + m_{1})
v_{2 }= 2mu_{1 } / ( m + m)
v_{2 }= u_{1}
3 ) when m_{1} = m_{2} = m ( say ) and u_{2} = 0
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} ) + 2m_{2}u_{2} / ( m_{1} + m_{2} )
v_{1 }= 0 + 0
v_{1 }= 0
and
v_{2 }= 2m_{1}u_{1 } / ( m_{2} + m_{1}) _{ }– u_{2}( m_{1} – m_{2} ) / ( m_{2} + m_{1})
v_{2 }= 2mu_{1 } / ( m + m)
v_{2 }= u_{1}
Practice questions
1 ) Define work.
Ans : When a force is applied on an object to displace it from its initial position then it is said to be work is done. It is the Dot Product of Force and Displacement.
2 ) what do you mean by positive work ?
Ans :
Positive Work : When the applied force and displacement of the objects take place in the same direction or at acute angle , then it is said to be positive work is done. That is the applied force and displacement are parallel to each other and angle between them is zero degree and the value of cos0 is 1 i.e
W = FS
OR
W = FS cosθ
Examples :
- Kicking of a football
- Moving a chair
- pushing and moving a table
- A car moving forward
- throwing a stone
3 ) what do you mean by negative work ?
Ans :
Negative Work : When the applied force and displacement of the objects take place in the opposite direction, then it is said to be negative work is done. That is the applied force and displacement are antiparallel to each other and angle between them is 180 degree and the value of cos180 is ( – 1 ) i.e
W = -FS
Examples :
- Two boys push each other
- In Tug of War when opponent team pulls hard.
- Trying to catch a heavy object, hands move down
4 ) what is meant by zero work ?
Ans :
Zero Work : When the applied force and displacement of the objects take place perpendicularly, then it is said to be zero work is done. That is the applied force and displacement are normal to each other and angle between them is 90 degree and the value of cos90 is 0 . when we apply a force on the object but it does not displace from its position, then there is zero work done .
W = 0
Examples :
- Pushing a wall
- Walking on the road
- Motion of planet on its orbit
5 ) Which physical quantity has [ ML^{2}T^{-2} ] as dimension ?
Ans : Work
6 ) What is the relation between work and displacement ?
Ans : Work ∝ displacement ( when force remains constant )
i.e greater is the displacement greater is the amount of work and vice versa.
7 ) What is the relation between work and velocity ?
Ans : Let
w = work
f = force
v = change in velocity
a = acceleration
m = mass of the object
d = displacement
w = fd = mad = mvd / t = mv ( d / t ) = mv^{2}
w = mv^{2}
w ∝ v^{2}
8 ) Write down the dimension of power.
Ans : [ ML^{2}T^{-3} ]
9 ) Derive the expression for kinetic energy of a body of mass ‘ m ‘ and velocity ‘ v ‘ .
Ans :
Let
u = initial velocity = 0 ( ∵ the object is initially at rest )
v = final velocity
a = acceleration
m = mass of the object
t = time
F = force applied on the object to displace it from its position
s = displacement
From Newton’s 2nd law of motion we have
F = ma —————————————— ( 1 )
From equation of motion we have
v^{2} = u^{2} + 2as
⇒ v^{2} = 2as
⇒ s = v^{2} / 2a
Now, workdone = force * displacement = ma * v^{2} / 2a = mv^{2} / 2
10 ) Calculate the potential energy in a string having a body of mass at one end ( this massive body is fixed at this end ) .
Ans :
Let us assume that a very light and perfectly elastic spring is fixed at one end from a rigid support. Other end of the support is attached with a body of mass ‘m’. ( In the above figure, red circular part is assumed as the body. Let the body is displaced from it’s equilibrium position. Due to elastic property of spring, the restoring force come into play and the whole system is seemed to be in to and fro motion.
Let
displacement be = x
restoring force = F
According to Hook’s Law
F ∝ – x —————-( 1 )
F = – kx
where k = spring constant or force constant
Work done against the restoring force is given by
w = ∫ Fdx
where dx = small displacement
w = ∫ ( – kx ) dx
w = – k ∫ xdx
w = – k x^{2} / 2 + c
where ‘c’ is integration constant
when x = 0 then w = 0
0 = – k ( 0 )^{2} / 2 + c
c = 0
w = – kx^{2} / 2
this work done is stored in the form of elastic potential energy given by
∴ U = kx^{2} / 2
11 ) Which principle of conservation is common in all types of collision ?
Ans : Principle of conservation of linear momentum
12 ) A body is at rest. Another body hits the stationary body with the velocity u. If both the bodies have equal masses then what will be the velocity of the stationary body after elastic collision ?
Ans :
Let
m_{1} = mass of first colliding object = m
m_{2} = mass of second colliding object = m
u_{1} = initial velocity of first colliding object = u
u_{2} = initial velocity of second colliding object = 0
v_{1} = final velocity of the first colliding object = v
v = final velocity of the second colliding object
applying principle of conservation of linear momentum
m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}
m_{1}u_{1} – m_{1}v_{1} = m_{2}v_{2} – m_{2}u_{2}
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} ) ———————————- ( 1 )
putting the value of u_{1} , v_{1,} m_{1}, m_{2} and u_{2}
m ( u – v ) = m ( v_{2} – 0 )
m ( u – v ) = v_{2 ————————————- ( 2 ) }
applying principle of conservation of kinetic energy
m_{1}u^{2}_{1} / 2 + m_{2}u^{2}_{2} / 2 = m_{1}v^{2}_{1} / 2 + m_{2}v^{2}_{2} / 2
m_{1}u^{2}_{1} – m_{1}v^{2}_{1} = m_{2}v^{2}_{2} – m_{2}u^{2}_{2}
m_{1} ( u^{2}_{1} – v^{2}_{1} ) = m_{2} ( v^{2}_{2} – u^{2}_{2} )
m_{1} ( u_{1} – v_{1} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
by using equation ( 1 ) we get
m_{2} ( v_{2} – u_{2} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
( u_{1} + v_{1} ) = ( v_{2} + u_{2} ) ——————————— ( 3 )
putting the value of u_{1} , v_{1} and u_{2}
u + v = v_{2} + 0
u + v = v_{2 ——————————- ( 4 ) }
On adding equation ( 2 ) and equation ( 4 ) we get
2 u = 2 v_{2}
u = v_{2}
hence the stationary body will move with the initial velocity of moving body i.e u after elastic collision.
13 ) What is the difference between elastic and inelastic collision ?
Ans : The difference between elastic and inelastic collision are here under.
Elastic collision |
Inelastic collision |
---|---|
1 ) Original shape of the colliding
bodies regain after elastic collision. |
1 ) Original shape of the collidingbodies do not regain after inelastic collision. |
2 ) There is no loss of kinetic energy
that is the kinetic energy is conserved. |
2 ) There is loss of kinetic energy that is the
kinetic energy is conserved. |
3 ) The value of coefficient of restitution
is unity. |
3 ) The value of coefficient of restitution
is not unity. |
4 ) Total energy of the colliding system
is constant. |
4 ) Total energy of the colliding system
is not constant. |
5 ) Mechanical energy of the system is not
converted into other form of energy like light energy, heat energy and sound energy. |
5 ) Mechanical energy of the system is not
converted into other form of energy like light energy, heat energy and sound energy. |
14 ) Explain conservative force and write down the it various properties ?
Ans : Conservative force : When the work done by a force in moving a body from one point to another is independent of the followed between the two points then the force is said to be conservative force.
Example : Gravitational force, elastic force and electrostatic force etc.
Properties of Conservative force :
- The work done is independent of path followed between the initial and final position of a body.
- The work done by this force depends only on the initial and final position of the body.
- The work done by this force in moving a particle along a closed path is zero.
- The work done by this force is always reversible.
- Mechanical energy of a body is always conserved under the action of conservative force.
15 ) What is non – conservative force ?
Ans : When the work done by a force in moving a body from one point to another is dependent of the followed between the two points then the force is said to be conservative force. It is also termed as dissipative force.
Example : Frictional force, force due to air resistance.
16 ) What is the difference between conservative and non conservative forces ?
Conservative force |
Non – conservative force |
---|---|
1 ) The work done is independent
of path followed between the initial and final position of a body. |
1 ) The work done is dependent
of path followed between the initial and final position of a body. |
2 ) The work done by this force
depends only on the initial and final position of the body. |
2 ) The work done by this force does not
depend only on the initial and final position of the body. |
3 ) The work done by this force in
moving a particle along a closed path is zero. |
3 ) The work done by this force in
moving a particle along a closed path is non zero. |
4 ) The work done by this force is
always reversible. |
4 ) The work done by this force is
always irreversible. |
5 ) Mechanical energy of a body is
always conserved under the action of conservative force. |
5 ) Mechanical energy of a body is
always conserved under the action of conservative force. |
17 ) What is power ? What is SI unit of power ? Prove that the power is scalar product of force and velocity.
Ans :
# Power : The rate of doing work is called as Power.
Power = Work / time
P = dw / dt
#SI – Unit of power is Watt.
# Let
F = force
s = displacement
w = work
v = velocity = s / t
t = time
P = power
Now Work = force . Displacement
w = F . s
here, the bold letters represent the vector notation.
P = w / t
P = F . s / t
P = F . v
Hence it is proved that power is the scalar product of force and velocity.
18 ) Derive the expression for gravitational potential energy of a body lying at a height ‘ h ‘ above the surface of the earth.
Ans :
Let an object of mass ‘ m ‘ is raised to a height of ‘ h ‘. The work done in raising the object is stored in the form of potential energy.
∴ Potential energy = Force by which the object is raised * height = Weight * height = mgh
where g = acceleration due to gravity.
P.E = mgh
19 ) Find the final velocities of the two bodies of different masses after the elastic collision in one dimension.
Ans :
Let
m_{1} = mass of first colliding object
m_{2} = mass of second colliding object
u_{1} = initial velocity of first colliding object
u_{2} = initial velocity of second colliding object
v_{1} = final velocity of the first colliding object
v = final velocity of the second colliding object
applying principle of conservation of linear momentum
m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}
m_{1}u_{1} – m_{1}v_{1} = m_{2}v_{2} – m_{2}u_{2}
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} ) ———————————- ( 1 )
applying principle of conservation of kinetic energy
m_{1}u^{2}_{1} / 2 + m_{2}u^{2}_{2} / 2 = m_{1}v^{2}_{1} / 2 + m_{2}v^{2}_{2} / 2
m_{1}u^{2}_{1} – m_{1}v^{2}_{1} = m_{2}v^{2}_{2} – m_{2}u^{2}_{2}
m_{1} ( u^{2}_{1} – v^{2}_{1} ) = m_{2} ( v^{2}_{2} – u^{2}_{2} )
m_{1} ( u_{1} – v_{1} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
by using equation ( 1 ) we get
m_{2} ( v_{2} – u_{2} )( u_{1} + v_{1} ) = m_{2} ( v_{2} – u_{2} )( v_{2} + u_{2} )
( u_{1} + v_{1} ) = ( v_{2} + u_{2} ) ——————————— ( 2 )
v_{2} = u_{1} + v_{1} – u_{2}
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} )
m_{1} ( u_{1} – v_{1} ) = m_{2} ( u_{1} + v_{1} – u_{2} – u_{2} )
( m_{1} u_{1} – m_{1}v_{1} )= ( m_{2}u_{1} + m_{2}v_{1} – 2m_{2}u_{2} )
( m_{2}v_{1} + m_{1}v_{1} ) = m_{1} u_{1} – m_{2}u_{1} + 2m_{2}u_{2}
v_{1}( m_{1} + m_{2} ) = u_{1}( m_{1} – m_{2} ) + 2m_{2}u_{2}
v_{1 }= u_{1}( m_{1} – m_{2} ) / ( m_{1} + m_{2} ) + 2m_{2}u_{2} / ( m_{1} + m_{2} )
from equation ( 2 ) we get
( u_{1} + v_{1} ) = ( v_{2} + u_{2} )
v_{1} = ( v_{2} + u_{2} – u_{1} ) ——————————— ( 3 )
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{1} ) = m_{2} ( v_{2} – u_{2} )
putting this value in equation ( 1 ) we get
m_{1} ( u_{1} – v_{2} – u_{2} + u_{1} ) = m_{2} ( v_{2} – u_{2} )
( m_{1}u_{1} – m_{1}v_{2} – m_{1}u_{2} + m_{1}u_{1} ) = ( m_{2}v_{2} – m_{2}u_{2} )
( m_{2}v_{2} + m_{1}v_{2} ) = ( m_{1}u_{1}– m_{1}u_{2} + m_{1}u_{1} + m_{2}u_{2} )
v_{2}( m_{2} + m_{1}) = ( 2m_{1}u_{1 }– m_{1}u_{2} + m_{2}u_{2} )
v_{2}= 2m_{1}u_{1 } / ( m_{2} + m_{1}) _{ }– u_{2}( m_{1} – m_{2} ) / ( m_{2} + m_{1})
20 ) What is the difference between kinetic energy and potential energy ?
21 ) Show that the mechanical energy of a body is conserved, when it falls freely under the action of gravitational field.
22 ) What is coefficient of restitution ? Discuss its value in different types of collision.
23 ) Discuss the elastic collision of two bodies of equal masses in two dimension.
24 ) Discuss the inelastic collision of two bodies of equal masses in two dimension.
25 ) when a body is at rest and other moving body hits it with w some initial velocity. The collision is perfectly inelastic collision in two dimension. Find the final velocity of the stationary body after the collision.