Hand-out (CBSE) Work – power and energy `
CLASS - XI 1.
PHYSICS By Prof. Amandeep Singh DATE: 18/07/2015
Work In Physics, work is said to be done whenever a force acts on a body and it displaces body parallel to its direction.
Let F = Constant force on the body S = Displacement of body in the direction of force Then mathematically W=FS When displacement of body is not in the direction of force, then W = Component of force in the direction of displacement x displacement
b)
When a spring is stretched, work done by the external (stretching) force is positive.
c)
While free fall, work done by gravity is +ive
When 𝜃 > 90° . It means force (or its component) is anti-parallel to displacement Cos is -ive Work = negative So when external force opposes the motion of body, then work done by that external force is – ive Example: a) While weightlifting work done by gravity is negative
W = (F cos𝜃) S = 𝐹⃗ . 𝑆⃗ W = 𝐹⃗ . 𝑆⃗ So work is a Scalar quantity
When force acting on body is not constant i.e. work done by variable force Consider variable force 𝐹⃗ displaces a body through small displacement
b)
When a spring is stretched, work done by the spring force is negative
c)
While pushing box on the floor, work done by friction is negative
ds (for small displacement ds , force can be assumed constant). ⃗⃗⃗⃗⃗ Small work done, dW = 𝐹⃗ . 𝑑𝑆 Total amount of work done will be obtained by integrating above equation from initial to final limit 𝑊 𝑠2 𝑑𝑠 ∫ dW = ∫ 𝐹⃗ . ⃗⃗⃗⃗⃗ 0
𝑠1
𝑠 W = ∫𝑠 2 𝐹⃗ . ⃗⃗⃗⃗⃗ 𝑑𝑠 1
W = Area under F – S curve
2.
When 𝜃 = 90° . It means force is perpendicular to the displacement of body
Nature of work done W = 𝐹⃗ . 𝑆⃗ = F S cos𝜃 When 𝜃 < 90°. It means force (or its component) is parallel to displacement
Cos 90° = 0 Work = 0 Example: a) When a coolie travels on a horizontal platform with a load on his head, work done by the coolie is zero. (When he walks on stairs, then W≠ 0) b) When a body moves in a circle the work done by the centripetal force is always zero because it acts perpendicular to the motion of body
Cos is +ive Work = +ive So when external force favours the motion of body, then work done by that external force is positive Example: a) While weightlifting work done by man’s lifting force is +ive
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Hand-out (CBSE) Work – power and energy 3.
Units of force Absolute units
Gravitational units
[SI ] = 1 Nm or joule (J) 1 J = 1Nm
[SI] : kg-F m 1 kg - m = (9.8 N)(1m) = 9.8 Nm = 9.8J
[C.G.S ] = dyn cm or erg (J) 1 erg = 1 dyn cm
[C.G.S] : gF cm 1 g - cm = (980 dyn) (1cm) = 980 dyn cm = 980 erg
A.
Bullet fired from gun can pierce through the target due to its kinetic energy Moving air can run wind mills due to their K.E Expression for kinetic energy Consider a body of mass 𝑚 is at rest. Let a constant force F acts on it and increases its speed from 0 to 𝑣 while displacing it by s 𝑣 2 − 𝑢2 = 2as 𝑣 2 = 2as because u = 0 a=
Relationship between J and erg
𝑣2
----- ①
2𝑠
Work done on by constant force F on particle W = F.S = (ma) s
1 J = 1Nm = (105 dyn) (100cm) = 107 dyn cm = 107 erg
𝑣2
W=𝑚(
2𝑠
1
) 𝑆 = 𝑚𝑣 2 2
This work done is measure of kinetic energy 1
K.E = 𝑚𝑣 2
Ques: Does work depend on frame of reference?
4.
2
CONSERVATIVE AND NON- CONSERVATIVE FORCES
Expression for kinetic energy when force is variable (Calculus method) Consider a body of mass 𝑚 is at rest. Let a variable force F acts on it and increases its speed from 0 to 𝑣 while displacing it by s
Conservative force Force said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and not on the path followed
Small amount of work done by variable force, dW = F ds 𝑑𝑣
dW = 𝑚 𝑎 ds = 𝑚
dt
𝑑𝑠 = 𝑚 𝑣 𝑑𝑣
To find total work done, integrate within limits 𝑣
W = 𝑚 ∫0 𝑣 𝑑𝑣 𝑣 𝑣2 | 2 0
W=𝑚| 1
W = 𝑚𝑣 2 2
This work is a measure of K.E of the system Let a force F moves a body from A to B via 3 paths. Force F is said to conservative when work done by force is independent of path taken to move body from A to B 𝑊𝐴−𝐵−1 = 𝑊𝐴−𝐵−2 = 𝑊𝐴−𝐵−3
1
K.E = 𝑚𝑣 2 2
B. Work energy principle It states that net work done by all the forces (internal +external) acting on the body = Change in K.E of the body.
Example of conservative forces are : Gravitational forces, Electrostatic forces, Elastic forces etc Ques: Prove that work done by or against conservative force in moving a body in closed path is zero. Ques: Prove that gravitational force is conservative in nature
(It is applicable only when whole of the work done has been used to increase K.E of the system only) W = ∆ K.E W = 𝐾𝑓 − 𝐾𝑖
Non- Conservative forces A force is said to be non-conservative if amount of work done by or against the force in moving a body from one position to another, depends on the path followed
Proof Consider a body of mass 𝑚 is moving with initial velocity u . Let a constant force F acts on it and increases its speed from 0 to 𝑣 while displacing it by s 𝑣 2 − 𝑢2 = 2as
Work done by or against Non-conservative forces in moving a body in any closed path is ≠ 0. Because there is some net loss of Energy.
a=
𝑣 2 −𝑢2 2𝑠
----- ①
Work done on by constant force F on particle W = F.S = (ma) s
All dissipative forces (i.e. which involve loss of energy in form of heat or sound) are non-conservative. Example: Force of friction, viscous Forces etc
𝑣 2 −𝑢2
W=𝑚( 1
2𝑠 2
1
W= 𝑚𝑣 - 𝑚𝑢 2
5.
6.
2
2
2
W = 𝐾𝑓 − 𝐾𝑖
MECHANICAL ENERGY It is energy associated with motion, position or configuration of an object It is of two types: Kinetic energy and potential energy Kinetic Energy It is energy possessed by the body due to its motion. Kinetic energy is measured in terms of work required to produce the motion or destroy the motion of body. Examples
1
) 𝑆 = 𝑚 (𝑣 2 − 𝑢2 )
C.
If work done on system by a force is +ive, then 𝐾𝑓 > 𝐾𝑖 i.e. Positive work increases K.E of system If work done on system by a force is -ive, then 𝐾𝑓 < 𝐾𝑖 i.e. negative work decreases K.E of system
Relationship between Kinetic energy and momentum
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Hand-out (CBSE) Work – power and energy 1
1
2
2𝑚
K = 𝑚𝑣 2 =
𝑚2 𝑣 2 =
PHYSICS By Prof. Amandeep Singh
𝑝2 2𝑚
p = √2 𝑚 𝐾 It's clear that a single particle cannot have K.E without having momentum and vice-versa Graphs NOTE: For system of particles Momentum is a vector quantity whereas kinetic energy is a scalar quantity. If the kinetic energy of a system is zero then linear momentum definitely will be zero but if net momentum of a system is zero then kinetic energy may or may not be zero. AIEEE 2003
7.
Restoring force, 𝐹𝑆 ∝ - 𝑥 𝐹𝑆 = −k𝑥 where K is called spring constant or FORCE constant Within elastic limit restoring force = Applied external force To find elastic potential energy is stretched of compressed spring Consider a massless spring having force constant K. Let this spring is displaced by 𝑥 by applying external force F. Potential energy stored in spring will equal to work done by external force F in stretching spring from x = 0 to x = 𝑥 against the restoring force
POTENTIAL ENERGY (U) P.E is the energy possessed by a body or a system by virtue of its position or configuration in a field of conservative force. Potential energy of a body at any point is defined as amount of work done in bringing the body from zero level to that point against the field without acceleration. Types of potential energies Gravitational P.E P.E associated with system due to separation between two bodies that attract each other via gravitational force P.E due gravity of earth U = 𝑚𝑔ℎ
Elastic P.E P.E associated with the system due to compression or extension of an elastic object
Since restoring force is variable (because it depends on displacement) Small amount of work required to displace spring by small displacement d 𝑥 dW = F d𝑥 --------①
1
We know within elastic limit magnitude of restoring force = magnitude of external force F So F = k𝑥 Put in equation --------① dW = k𝑥 d𝑥
U = K 𝑥 2 or 2 1
2
2
U = K ( 𝑥2 - 𝑥1 ) 2
A.
Gravitational potential energy (due to earth gravity)
To find total work done, integrate above equation within limits 𝑥
W = k ∫0 𝑥 𝑑𝑥 =
1 2
K 𝑥2
This work done in displacing spring from x = 0 to x = 𝑥 gives measure of P.E stored in spring. 1
U = K 𝑥2 2
Consider a body of mass 𝑚 at height ℎ above the earth’s surface. To find gravitational P.E associated with the body, we have to find work required to bring the body from zero level to that height Let ground is taken as zero level. At any point weight mg of body acts downwards, so equal amount of force is required in upward direction to take the body from ground to height ℎ against the field without accelerating. W = F S = mgh This work done is a measure of gravitational potential energy of the earth-body system U = mgh
B.
Ques: Show that potential energy stored in spring is given by area under F-𝑥 curve
8.
CONSERVATION OF MECHANICAL ENERGY For a body in presence of conservative forces, the sum of kinetic and potential energies at any point remains constant throughout the motion. This is known as the law of conservation of mechanical energy. K + U = constant Note: It is not valid if non-conservative forces (e.g. friction) are also acting. Show that law of conservation of mechanical energy is valid in case of freely falling body
Elastic Potential energy Whenever spring is stretched or compressed, a restoring force comes into picture to bring it to the normal position. This restoring force obey Hook’s law According to Hooke’s law this restoring force is proportional to the displacement 𝑥 and Its direction is always opposite to the displacement.
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Hand-out (CBSE) Work – power and energy
11. COLLISIONS A collision is said to occur between two bodies, either if they physically collide against each other or if path of one is affected by the force exerted by other
At Point A
At Point B
At Point C
𝑣 = 0 K.E = 0 U = 𝑚𝑔ℎ T.E = 𝑚𝑔ℎ
𝑣 2 - 𝑢2 = 2 g 𝑠 u=0 𝑣2 = 2 g 𝑥 1 1 𝐾𝐵 = m 𝑣 2 = m(2g𝑥) = 𝑚𝑔𝑥
𝑣 2 - 𝑢2 = 2 g 𝑠 u = 0 and s = ℎ 𝑣2 = 2 g ℎ
2
2
P.E = 𝑚𝑔 (H-𝑥) T.E = 𝑚𝑔𝑥 + mg (H-𝑥) = 𝑚𝑔H
K.E = 𝑚𝑔ℎ P.E = 0 T.E = 𝑚𝑔ℎ
b)
NOTE: a) For collision to occur, actual physical contact is not necessary. e.g. In Rutherford's experiment, 𝛼 - particles get scattered due to electrostatic repulsion between 𝛼 - particles and nucleus from the distance During collisions we ignore external forces like friction and gravity (because during collision impulsive forces are much larger than external forces). So momentum remains conserved Perfectly Elastic collision If during collision there is no loss of K.E it is called Perfectly elastic collision
So clearly total mechanical energy of body is always constant during free fall. During free fall potential energy is continuously being converted into kinetic energy of the body. Ques: Draw variation of P.E and K.E of body during free fall Ques: Is mechanical energy also conserved when spring is stretched and released, assuming no friction present. Also find K.E and P.E of spring at different positions.
9.
Power Power of a body is defined as the rate at which the body can do the work or work is done on the body. Average power ( 𝑃𝑎𝑣𝑔 ) =
𝑊𝑜𝑟𝑘
P=
⃗⃗⃗⃗ F . ⃗S⃗ t
dt
⃗⃗⃗ . 𝑣⃗ = F
[SI units]: J/s or Watt [Practical unit]: horse power (hp) = 746 W
e=1
10. Einstein’s mass-energy equivalence Einstein discovered that mass can be converted into energy and viceversa. E = m 𝑐 2 where c = speed of light If mass 'm' disappears, energy E = m 𝑐 2 appears If Energy E disappears , mass m =
If during collision there is some loss of K.E , it is called Perfectly inelastic collision Some energy is lost as heat etc. Total momentum of system is conserved Total energy is conserved K.E is not conserved Part of K.E is converted into heat, sound energies etc. Non-conservative forces are also involved along with conservative forces e.g. Collision between car and bus Collisions taking place in daily lives
0
Perfectly inelastic collision When two bodies stick together after collision and move as a single body with common velocity, this type of collision is called Perfectly inelastic collision Total momentum of system is conserved Total energy is conserved Loss of K.E is maximum
e.g. Bullet fired into wooden block and bullet remain embedded in it and both moves together e=0
There is one more type of collision called super-elastic collision or explosive collisions: During the collision there is increase in K.E. This occurs if there is release of P.E on an impact.
[P] = [𝑀1 𝐿2 𝑇 −3 ]
Only conservative forces are involved during collision E.g. Collisions between sub atomic particles
𝑡𝑖𝑚𝑒 dW
Instantaneous power, P =
Total momentum of system is conserved Total energy is conserved K.E is conserved
Inelastic collision
𝐸 𝑐2
12. 1-D Elastic collision 1-D collision means before and after collision particles moves in a straight line
appears
Examples a) Annihilation of matter Electron and Positron comes close to each other, they annihilate (destroy) each other producing energy (2𝛾 gamma ray photons) (-e) + (+e) -----> 2𝛾 rays b)
Energy generation in stars and sun is due to the conversion of mass into energy
Principle of conservation of energy It states that total energy + total mass of the universe remains constant. If one part of universe loses energy or mass, other part must gain equal amount of energy or mass
Consider two elastic bodies having masses 𝑚1 , 𝑚2 moving in straight line with initial velocities 𝑢1 , 𝑢2 collide head on elastically . Let after collision their velocities become 𝑣1 and 𝑣2 . Linear Momentum is conserved
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Hand-out (CBSE) Work – power and energy 𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2 ---------- ① 𝑚1 (𝑢1 - 𝑣1 ) = 𝑚2 (𝑣2 - 𝑢2 ) ----------- ②
𝑢2 = 0
Kinetic Energy is conserved 1 2
1
1
1
2
2
2
𝑚1 𝑢1 2 + 𝑚2 𝑢2 2 = 𝑚1 𝑣1 2 +
𝑚2 𝑣2 2
𝑚1 (𝑢1 - 𝑣1 ) (𝑢1 + 𝑣1 ) = 𝑚2 (𝑣2 - 𝑢2 ) (𝑣2 + 𝑢2 ) ----------③ Dividing equation ③ by ② 𝑢1 + 𝑣1 = 𝑣2 + 𝑢2 𝑢1 - 𝑢2 = 𝑣2 - 𝑣1 ------④ 𝑢1 - 𝑢2 = 𝑣2 - 𝑣1 Relative velocity of approach = Relative velocity of separation Coefficient of restitution (e) It is defined as ratio of relative velocity of approach to relative velocity of separation
A.
e=
𝑣2 − 𝑣1
Loss in K.E =
1
𝑚1 𝑚2
2 m1 +m2
% age loss in K.E of 𝑚1 =
% age loss in K.E of 𝑚1 =
(This K.E is lost by 𝑚1 only)
𝐾.𝐸 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑚1 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐾.𝐸 𝑜𝑓 𝑚1
% age loss in K.E of 𝑚1 =
𝑣2 − 𝑣1
For 1-D elastic collision, e = 1 For perfectly elastic collisions, e = 0 Because bodies stick to each other after collision. i.e. relative velocity of separation is zero in that case.
(𝑢1 )2
x 100
1 𝑚1 𝑚2 (𝑢1 )2 2 m1 +m2 1 𝑚1 (𝑢1 )2 2
𝑚2 m1 +m2
x 100
x 100
%
To find expression of final velocity 𝑣1
B.
From equation ④ 𝑣2 = 𝑢1 - 𝑢2 + 𝑣1 Put in equation ① 𝑣1 = (
m1 −m2 m1 +m2
) 𝑢1 +
2𝑚2 m1 +m2
𝑢2 --------⑦
To find expression of final velocity 𝑣2
C.
From equation ④ 𝑣1 = 𝑢1 - 𝑢2 + 𝑣2 Put in equation ① 𝑣2 = (
m2 −m1 m1 +m2
) 𝑢2 +
2𝑚1 m1 +m2
𝑢1 ----------⑧
Special cases 13. 1-D perfectly inelastic collision Consider two elastic bodies having masses 𝑚1 , 𝑚2 moving in straight line with initial velocities 𝑢1 , 𝑢2 collide head on . Let after collision two bodies stick to each other and move with common velocity 𝑣 Linear Momentum is conserved 𝑚1 𝑢1 + 𝑚2 𝑢2 = ( 𝑚1 +𝑚2 ) 𝑣 K.E before collision 1 2
1
𝑚1 𝑢1 2 + 𝑚2 𝑢2 2 2
K.E after collision 1 2
( 𝑚1 + 𝑚2 )𝑣 2 1
1
1
2 2 1 𝑚1 𝑚2
2
Loss in K.E = ( 𝑚1 𝑢1 2 + 𝑚2 𝑢2 2 ) - ( 𝑚1 + 𝑚2 )𝑣 2 Loss in K.E =
2 m1 +m2
(𝑢1 − 𝑢2 )2
Special case When target (𝑚2 ) is at rest and body of mass 𝑚1 collide and stick to it
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