Work Power Energy

WORK, ENERGY, POWER & MOMENTUM WORK Work is said to be done by a force when the point of application of the force is displaced in the direction of the force. Work is a scalar quantity and is measured by the product of the magnitude of the force and the component of displacement along the direction of the applied force. Constant

Force

Variable

WORK DONE BY A CONSTANT FORCE If the direction and magnitude of a force applying on a body is constant, the force is said to be constant. Work done by a constant force, W = Force × component of displacement along force = displacement × component of force along displacement. 



If a force F is acting on a body at an angle to the horizontal and the displacement r is along the horizontal, F the work done will, be W = (F cos ) r Fcos = F (r cos )   

r

In vector from, W = F . r 



If F  ˆi Fx  ˆj Fy  kˆFz and r  ˆi x  ˆj y  kˆz , the work done will be, W = Fx x + Fy y + Fzz Note : The force of gravity is the example of constant force, hence work done by it is the example of work done by a constant force.

WORK DONE BY A VARIABLE FORCE   The equation W  F.S  FScos  is valid when F remains constant but when the force is variable, total   work done is obtained by integration method using W   dW   F.dS. 1

Physics : Work, Energy, Power and Momentum

(A)

WHEN THE FORCE IS SOME FUNCTION OF POSITION Let a force F  F(x) act along the x-axis as shown in the F–x graph in the figure. Work done during small displacement dx is F dx which is equal to the area of the shaded rectangular strip. F(x) Hence the total work done is B

F   F dx  Area under the F –x graph. A

An example of the variable force is the spring force in which the force F is proportional to the extension x, given by F = – kx, where k = force constant and negative sign indicates opposite direction of spring force F and extension x.

A

f   1 2 2  Work done by spring =  F.dx    kx dx   k(x f  xi ) . 2 x1

dx

B

x

x

If spring be initially relaxed, then work done by the spring (or, work done against the external agent) for an extension x is obtained by putting xi  0 and x f  x leading to

1 W   kx 2 . 2 Note that the work done by the spring is negative where as the work done by the external agent producing the extension is positive. The two, however, are equal in quantity. (B)

WHEN THE FORCE IS TIME DEPENDENT: Let

F  F(t) , then

     dx   W   F.dx   F.  dt   F.vdt ,  dt      where F and v are time dependent instantaneous force and velocity vectors expressed as functions of time.

CALCULATION OF WORK DONE FROM FORCE DISPLACEMENT GRAPH 

Suppose a body, whose initial position is r1, is acted upon by a variable force F and consequently the body acquires its final position r2. From position r to r + dr or for small displacement dr, the work done will be Fdr whose value will be the area of the shaded strip of width dr. The work done on the body in displacing it from position r1 to r2 will be equal to the sum of areas of all the such strips r2

Thus, total work done,

W =

 dW

P2 F

r1

r2

=

 F.dr

O

P1 M r1

r1

= Area of P1P2NM The area between the graph between force and displacement axis is equal to the work done. 2

dr

N r2

Physics : Work, Energy, Power and Momentum

Note : To calculate the work done by graphical method, for the sake of simplicity, here we have assumed the direction of force and displacement as same, but if they are not in same direction, the graph must be plotted between F cos and r. (i)

Work is a scalar quantity

(ii)

The dimensions of work : ML2T–2

(iii)

Unit of work : there are two types of unit of work (a)

Absolute unit : Joule (in M.K.S), Erg (in C.G.S.) (Note : 107 erg = 1 joule)

(b)

Gravitational unit : Kilogram - metre (in M.K.S), Gram-cm (in C.G.S)

(Note : 1 kilogram metre = 9.8 joule = 105 gram cm)

NATURE OF WORK DONE Although work done is a scalar quantity, yet its value may be positive, negative or even zero  

(a)

Positive work : As W = F . r  F r cos  When is acute (90º), cos is negative. Hence work done is negative

(i)

Examples : When a body is thrown up, its motion is opposed by gravity. The angle  between the gravitational 



force F and displacement r is 180º. As cos = – 1, therefore, work done by gravity is negative. (ii)

When a body is moved over a rough horizontal surface, the motion is opposed by the force of friction. Hence work done by frictional force is negative. Note that work done by the applied force is not negative

(iii)

(c)

When a positive charge is moved closer to another positive charge, work done by electrostatic force of repulsion between the charges is negative. 



Zero work : When force F or the displacement r or both are zero, work done W, will be 



zero. Again when angle between F and r is 90º, the work done will be zero. Examples : (i) (ii) (iii)



When we fail to move a heavy stone, however hard we may try, work done by us is zero, r  0 When a collie carrying some load on his head moves on horizontal platform, = 90º. Therefore, workdone by the collie is zero. This is because = 90º Tension in the string of simple pendulum is always perpendicular to displacement of the bob. Therefore, work done by tension is always zero. 3

Physics : Work, Energy, Power and Momentum

Note : Another way of expressing negative or positive work is that when energy is transferred to the object work done is positive and when energy is transferred from the object the work done is negative and hence the work which is a transfer of energy has same dimensions as energy.

WORK DEPENDS ON THE REFERENCE FRAME Consider a situation in which a person pushes a block inside a moving train through a distance S. The work   done on the block in the train’s frame will be F.S . The work done on the block in the ground frame     will be F.(S  S0 ) , where S0 is the displacement of the train relative to the ground. In mechanics, we encounter two types of forces : conservative and non-conservative force.

CONSERVATIVE AND NON-CONSERVATIVE FORCES Conservative Force: A force is said to be conservative if the work by the force in moving a mass through any closed path is zero. Alternative, if the net work done by the force in moving a mass between two points is path independent i.e., the work done depends only on the location of two points and not on the path followed, the force may be termed conservative.Conservative forces are non-dissipative and can store work (as energy) Examples Gravitational Force, electrostatic force, ideal spring force are examples of conservative forces.Associated with each conservative force there exists a well defined function known as the potential energy function. In these three cases, we refer to the respective energies as gravitational potential energy, electrostatic potential energy and strain energy.

Non-conservative Force The forces which do not fulfill the above mentioned conditions are non-conservative. For these the energy isnot stored but dissipated during displacement. Examples of non-conservative forces are friction, viscous forces etc.

EXAMPLE BASED ON WORK 

Example 1.

A position dependent force F  7  2 x  3 x 2 acts on a small body of mass 2kg and displaces it from x = 0 to x = 5 m. The work done in joule will be

Solution :

W

X2



X1

Example 2.

5

Fdx 

 0

5

 2x 2 3x 3  (7  2x  3x 2 )dx  7x     135J 2 3 0 

For the force displacement diagram shown in adjoining diagram the work done by the force in displacing the body from x = 1 cm to x = 5 cm is 20 10 F 0 (In dyne) -10 -20

Solution : 4

1 2 3 4

5 6 7 8

x-(cm)

Work = Area under the curve and displacement axis = 10 + 20 – 20 + 10 = 20 erg.

Physics : Work, Energy, Power and Momentum

Example 3.

A uniform chain of mass M and length L is lying on a frictionless table in such a way that its 1/ 3 part is hanging vertically down. The work done in pulling the chain up the table is

Solution :

If length x of the chain is pulled up on the table, then the length of hanging part of the chain ML L   would be   x  and its weight would be   x  g. If it is pulled up further by a distance 3 L 3     dx, the work done in pulling up.

j 

ML    x  g dx L 3  L/3



w

 0

ML MgL    x  g dx  L 3 18 

Example 4.

The work done in pulling a body of mass 5 kg along an inclined plane (angle 60º) with coefficient of friction 0.2 through 2m, will be

Solution :

The minimum force with a body is to be pulled up along the inclined plane is mg (sin + cos)   Work done, W  F. d = Fd cos 0º = mg (sin +  cos ) × d = 5 × 9.8 (sin 60º + 0.2 cos 60º) × 2 = 98.08 J

Example 5.

  A particle moves from a point r1  (5m)iˆ  (5m)ˆj to another point r2  (6m)iˆ  (4m)ˆj during  which a certain force F  (5N)iˆ  (5N)ˆj acts on it. Find the work done by the force on the particle during the displacement.

Solution :

Displacement vector    S  r2  r1 = ˆi  ˆj m .    Work done = F.S  (5iˆ  5ˆj).(iˆ  j )

 

= (5 – 5) = 0.

ENERGY The energy of a body is defined as the capacity of doing work. There are various form of energy (i)

mechanical energy

(ii) chemical energy

(iii)

electrical energy

(iv) magnetic energy

(v)

nuclear energy

(vi) sound energy

(vii)

light energy etc

Energy of system always remain constant it can neither be created nor it can be destroyed, however, it may Example be converted from one form to another

5

Physics : Work, Energy, Power and Momentum

Motor

Mechanical energy



Generator

Electrical energy

  

Photocell

Electrical energy

Electrical energy

Heater   

Heat energy

Electrical energy

Radio  

Sound energy

Nuclear energy



Chemical energy



Electrical energy



Heat energy



Electrical energy

  

Mechanical energy Light energy

Nuclear Re actor

Electrical energy

Cell

Electrical energy

Secondary Cell

Chemical energy

Incendence nt lamp

Light energy

Energy is a scalar quantity Unit : Its unit is same as that of work . In MKS : Joule, watt sec In CGS : Erg Note : 1 eV = 1.6 × 10–19 joule 1 KWh = 36 × 105 joule 107 erg = 1 joule Dimension [M1L2T–2] According to Einstein’s mass energy equivalence principle mass and energy are inter convertible i.e. they can be changed into each other Energy equivalent of mass m is, E = mc2 Where,

m : mass of the particle c : velocity of light E : equivalent energy corresponding to mass m.

In mechanics we are concerned with mechanical energy only which is of two type (a) kinetic energy; (b) potential energy

KINETIC ENERGY The energy possessed by a body by virtue of its motion is called kinetic energy If a body of mass m is moving with velocity v, its kinetic energy

6

Physics : Work, Energy, Power and Momentum

KE =

1 mv2, for translatory motion 2

KE =

1 I2, for rotational motion 2

Kinetic energy is always positive If linear momentum of body is p, KE =

p2 1  mv 2 - for translatory motion 2m 2

If angular momentum of body is J, KE =

J2 1 2  I - for rotational motion 2I 2

p or J  E

p : momentum

E

E

P

1/P

The kinetic energy of a moving body is measured by the amount of work which has been done in bringing the body from the rest position to its present moving position or The kinetic energy of a moving body is measured by the amount of work which the body can do against the external forces before it comes to rest. If a body performs translatory and rotational motion simultaneously, its total kinetic energy =

1 1 mv 2  I2 2 2

WORK ENERGY THEOREM For translatory motion : Work done by all the external forces acting on a body is equal to change in its kinetic energy of translation. Work done by all the external forces = change in K.E of translation =

1 1 mv 22  mv 12 2 2

For rotational motion : Work done by all the external torque acting on a rigid body is equal to change in its 1 2 1 2 rotational kinetic energy. Work done by all the external torque = I2  I1 2 2 Note : In simple words K = Kf – Ki = W in the work energy theorem ?

POTENTIAL ENERGY The energy which a body has by virtue of its position or configuration in a conservative force field Potential energy is a relative quantity. Potential energy is defined only for conservative force field. Potential energy of a body at any position in a conservative force field is defined as the work done by an external agent against the action of conservative force in order to shift it from reference point. (PE = 0) to the present position or.

7

Physics : Work, Energy, Power and Momentum

Potential energy of a body in a conservative force field is equal to the work done by the body in moving from its present position to reference position.   U ˆ U ˆ U ˆ i j Relationship between conservative force field and potential energy (U) F U  grad (U) = – k x y z

Example 6.

(i)



U = 3x2  F  6 x ˆi

(ii) U = 2x2y + 3y2x + xz2



 F  ( 4 xy  3 y 2  z 2 )  (2x 2  6 xy )  (2xz) dU If force varies only with one dimension, then F = – or dx

x2

u2

d

U

u1

 Fdx

x1

Potential energy may be positive or negative (i)

Potential energy is positive, if force field is repulsive in nature

(ii)

Potential energy is negative, if force field is attractive in nature

If r  (separation between body and force centre), U , force field is attractive or vice-versa.

Repulsion forces U+ve U-ve

r Attraction forces

If r , U , force field is repulsive in nature. DIFFERENT TYPES OF FORCES AND CORRESPONDING P.E (a)

Gravitational potential energy : For small distances above or below the earth surface : Reference point = Earth surface i.e. P.Esurface = 0 P.E above the earth surface is positive and below the earth surface is negative and in magnitude it is equal to mgh. For greater distance : Reference point = i.e. P.E = 0 P.E. = –

GMm , for r > R r

Where R = radius of earth, r = distance of body from the centre of earth, m = mass of body, M = mass of earth Electrostatic potential energy : Reference point = i.e. P.E = 0 P.E. = –

KQ1Q 2 , [value of Q1 and Q2 are substituted with sign.] r

Intermolecular potential energy : Reference point = i.e. P.E = 0  dU 6b  2a / b a b    1 F=– U (r) = 12  6 , 7  6 dr r  r  r r Elastic potential energy :

8

Physics : Work, Energy, Power and Momentum

Which is associated with the state of compression of extension of an elastic object U (x) = k = spring constant, x = change in dimensions

1 2 kx where 2

POTENTIAL ENERGY CURVE A graph plotted between the PE of a particle and its displacement from the centre of force field is called PE curve. Using graph, we can predict the rate of motion of a particle at various positions. Force on the particle is F(x) = –

dU dx

Case : I On increasing x, if U increase, force is in (–)ve x-direction i.e. attraction force. Case : II On increasing x, if U decreases, force is in (+)ve x-direction i.e. repulsion force. Different positions of a particle Position of equilibrium If net force acting on a body is zero, it is said to be in equilibrium

dU = 0 Points P, Q, R and S are the states dx

of equilibrium positions. TYPES OF EQUILIBRIUM Stable equilibrium - When a particle is displaced slightly from a position and a force acting on it brings it back to the initial position, it is said to be in stable equilibriums position. Necessary conditions 

d2U dx 2

  ve

Unstable equilibrium : When a particle is displaced slightly from a position and force acting on it tries to displace the practice further away from the equilibrium position, it is said to be in unstable equilibrium. Condition :



dU = 0, dx

dU d2U   ve = 0 potential energy is max i.e. = dx dx 2

Neutral equilibrium : In the neutral equilibrium when a particle is displaced from its position it does not experiences any force to acting on it and continues to be in equilibrium in the displaced position, it is said to be in neutral equilibrium.

EXAMPLE BASED ON ENERGY Example 7.

A meter scale of mass m initially vertical is dispalced at 45º keeping the upper end fixed, the change in PE will be-

Solution :

Work = change in PE = Force × displacement

 dU = mg (1  cos ) 2 1 º = mg × (1  cos 45 ) 2



mg (  = 1m) = 2

 1  1   2 

/2 45º

G'

G

9

Physics : Work, Energy, Power and Momentum

Example 8.

If the potential energy function for a particle is U = a – oscillation will be.

b c  x x2

.........(1)



dU b 2c  2  3 dx x x

.........(2)

and

d2U 1  dx 2 x 3

Solution :

`

b c  ; The force constant for x x2

U=a–

6c    2b  x   

.........(3)

for equilibrium dU 0 dx



Solution :

Substituting this value in (3)

d2U

 b    2  2c  dx

3

6c  b 4  2 b     2c / b  8c 3 

d2U

K K = b4/8c3 dx 2 On passing through a wooden sheet a bullet looses 1/20 of initial velocity. The minimum number of sheets required to completely stop the bullet will beas

Example 9

2c b

x =

Use v2 = u2 + 2as for a sheet of thickness s v = (19/20)u

 19  2  20 u   u  2as   2as = (361/400)u2 – u2 suppose for n sheet v = 0



n=–

u2 = 2as

u2  39  u 2 2 s   400  2s

 39  u 2  a=–   400  2s 02 = u2 + 2a (ns)  11

Example 10.

The work done in taking out 2 litre of water using a bucket of mass 0.5 kg from a well of depth 6m will be-

Solution :

W = mgh = (mbucket + mwater)gh

[2 Lit water = 2 kg water]

= (0.5 + 2.00) × 9.8 × 6 = 15 × 9.8 = 147 J Example 11.

A body has velocity 200 m/s and its kinetic energy is 200 J. The mass of the body would be

1 mv 2  E 2

Solution :

Example 12. 10

or

 2E  2  200 4  102 1 m 2   2 4 100 4  10  v  (200 )



m = 0.01 kg

A body of mass 8 kg moves under the influence of a force. The position of the body and time are related as x = t2/2 where x is in meter and t in sec. The work done by the force in first two

Physics : Work, Energy, Power and Momentum

seconds. Solution :

Work done = change in kinetic energy 2

or

1 1  dx  1  2t  mv 2  m   m  2 2  dt  2 2

2

2

1 2 2 8    16Joules 2  2 

 Example 13.

A body falls on the surface of the earth from a height of 20 cm. If after colliding with the earth, its mechanical energy is lost by 75%, then body would reach upto a height of

1 mgh  mgh' 4

Solution :  Example 14.

h' 

Potential energy function describing the interaction between two atoms of a diatomic molecule is U( x ) 

Solution :

h 1   20  5cm 4 4 a



b

. In stable equilibrium, the distance between them would be x x6 In stable equilibrium potential energy is minimum. For minimum value of U(x)

or or 

12

d [U( x )]  0 dx d  a b   12  6   0 dx  x x  6 ( 2a  bx 6 )  0 x 13

 2a  x   b 

12a

or or

x

13



6b x7

0

bx6 – 2a = 0

1/ 6

Example 15.

Two electrons are at a distance of 1 × 10–12m from each other. Potential energy (in eV) of this system would be

Solution :

Potential energy of the system

U  

Kq1q2 r 9  10 9  1.6  10 19  1.6  10 19 1 10 12 23.04  10 17 1.6  10 19

= 23.04 × 10—17 Joule

eV  1.44  10 3 eV

Example 16.

The stopping distance for a vehicle of mass M moving with speed v along level road, will be ( is the coefficient of friction between tyres and the road)

Solution :

When the vehicle of mass m is moving with velocity v, the kinetic energy of the where K = 1/2 mv2 and if S is the stopping distance, work done by the firction W = FS cos  = m MgS cos 180º = – m MgS So by Work-Energy theorem, W =  K = Kf – ki 

– MgS = 0 – 1/2 Mv2 11

Physics : Work, Energy, Power and Momentum



S=

v2 2g

Example 17.

A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to (–k/r2), where k is constant. The total energy of the particle is

Solution :

As the particle is moving in a circle, so mv 2 k  2 r r F

Now as

1 k mv2 = 2 2r

Now

K.E =



P.E, U   F dr

r

dU dr





r

k     2  dr r  





k r

So total energy = U + K.E.

k k k   r 2r 2r Negative energy means that particle is in bound state. 

Example 18.

The work done by a person in carrying a box of mass 10 kg. through a vertical height of 10 m is 4900J. The mass of the person is

Solution :

Let the mass of the person is m . Work done, W = P.E at height h above the earth surface. = (M + m) gh or

4900 = (M + 10) 9.8 x 10

or

M = 40 kg

Example 19.

A uniform rod of length 4m and mass 20kg is lying horizontal on the ground. The work done in keeping it vertical with one of its ends touching the ground, will be -

Solution :

As the rod is kept in vertical position the shift in the centre of gravity is equal to the half the length =  /2 Work done

w = mgh = mg = 20 x 9.8 x

 2 4 = 392 J 2

Example 20.

A man throws the bricks to the height of 12 m where they reach with a speed of 12 m/sec. If he throws the bricks such that they just reach this height, what percentage of energy will he save.

Solution :

In first case,

W

1

=

1 m(v1)2 + mgh 2



1 m(12)2 + m × 10 × 12 2

= 72 m + 120 m

12

Physics : Work, Energy, Power and Momentum

and in second case,

W2 = mgh = 120 m

The percentage of energy saved 

192m  120m  100  38% 192m

POWER (a)

The time rate of doing work is called power

(b)

dw  d x   Power =  F.  F.v dt dt



 In translatory motion : P   F.v  

In rotational motion : P   .  (c)

It is scalar quantity

(d)

Unit : In MKS - J/sec, watt In CGS - erg/sec, (Note : 1KW = 103 watt, 1 HP = 746 watt)

(e)

Dimension : [M1L2T–3] Note : Power is the rate at which applied force transfers energy w where w work is done in t time t

(a)

Power P 

(b)

Instantaneous power P =

dw . dt

EXAMPLE BASED ON POWER Example 21.

A pump can lift 9000 kg coal per hour from a mine of depth 120 m. Assuming its efficiency is 75%, its power will be (in watts) -

Solution :

Power =

work time

ouput power 3 output power Effeiciency of pump = input power ; = 4 P Output power = 

3P 4

mgh 3 P = t 4

P =

4 3

mgh t

=

4 9000  9.8  120  = 3920 W 3 3600

13

Physics : Work, Energy, Power and Momentum

Example 22.

Solution :

A person of mass 60 kg is capable of taking a 15 kg massive objets to a height of 10 m in 3 min. The efficiency of person is -

work output mgh % Efficiency   work input  100 = (m  M)gh  100  m   15    100 =    100 = 20% =  m M  15  60 

Example 23.

An automobile of mass m accelerates from rest. If the engine supplies constant power p, the velocity at time t is given by

Solution :

Given that power = Fv = p = constant dv vp dt

or

m

or

 v dv   m dt

[as F = ma =

mdv ] dt

p

v2 p  t  C1 2 m Now as initially, the body is at rest i.e. v = 0 at t = 0 so, C1 = 0 

v

2pt m

Example 24.

In the above problem, the position (s) at time (t) is given by

Solution :

By definition v 





ds dt

 2pt  ds     m 



or

ds  2pt    dt  m 



 2p  s  m



 8p  s    9m 

1/ 2

1/ 2

1/ 2

dt

Now as t = 0, s = 0, so C2 = 0

[From (1)]

2 3/2 t  C2 3

1/ 2

t3 / 2

LINEAR MOMENTUM  The product of mass and velocity of the body is called the momentum  P mv

Momentum is a vector quantity and its direction is always along the direction of velocity. Unit : In MKS : kg-m/sec or N-sec In C.G.S gm-cm/sec or Dyne-sec Dimensions : [M1L1T–1] The rate of change of momentum of a body is equal to the magnitude of applied external force 

= Fext 

14



dP for v > m2

(c)

m1 m1). The fraction of initial kinetic energy converted into heat would be

Solution :

From momentum conservation law m1v1 + 0 = (m1 + m2) v 

v

m1v 1 (m1  m 2 )

Total kinetic energy before collision  Total kinetic energy after collision 





 m1v 1 1 (m1  m 2 )  2  m1  m 2

   

2



1 1 m1v 12  0  m1v 12 2 2

1 (m1  m 2 ) v 2 2

1 m12 v 12 2 (m1  m 2 )

Loss of kinetic energy



1 1 m12 v 12 1 m1v 12 (m1  m2  m1 ) m1v 12   2 2 (m1  m2 ) 2 (m1  m2 )



1 m1m2 v 12 2 (m1  m2 )

Fraction of initial kinetic energy converted into heat.



m2 1 m1m 2 v12 1 / m1v 12  m1  m 2 2 (m1  m 2 ) 2

Example 15.

A block of mass m 1 = 150 kg is at rest on a very long frictionless table, one end of which is terminated in a wall. Another block of mass m2 is placed between the first block and the wall, and set in motion towards m1 with constant speed u2. Assuming that all collisions are completely elastic, find the value of m2 for which both blocks move with the same velocity after m2 has collided once with m1 and once with the wall. (The wall has effectively infinite mass.)

Solution :

Let after the collision, v1 = speed of mass m1 towards left v2 = speed of mass m2 towards right. Hence, momentum before collision = momentum after collision m2u2=m1v1 – m2v2

.....(1)

The mass m2 rebounds elastically from the wall and its speed its reversed after the collision with the wall. 36

Physics : Work, Energy, Power and Momentum

According to the problem, the mass m2 has the same speed as that of mass m1 after its collision with the wall i.e. v2 = v1 . From eq. m2u2 = (m1 – m2) v1

.....(2)

Since the collision is elastic, then

1 1 1 m 2u 22  m1v12  m 2 v12 2 2 2

m2u22 = (m1 + m2) v12 From eq. (2), v 1 

......(3)

m 2u 2 m1  m 2

Substituting this value of v1 in eq. (3), we get

m 2u22 

(m1  m 2 )(m2u2 )2 (m1  m2 )2

or

(m1 – m2)2 = (m1 + m2) (m2)

or

m12 + m22 – 2m1 m2 = m1 m2 + m22

or

m12 = 3 m1 m2 m1 = 3m2

or

m2 

m1 150   50.kg 3 3

POINTS TO REMEMBER dW  tan  dt



The slope of work-time curve at any instant gives us the power at that instant, as P 



Area under power-time curves gives us total work done by/on the body, as W   P dt



When a man rowing a boat unstream is at rest with respect to shore, he is doing no work, though he is applying a force. However, when he stops rowing and moves down with the stream, work is done on him.



Power is dissipated only by the tangential component of force and not by the normal/radial component of force. Thus power dissipated by centripetal force is zero.



Work done against friction on a rough horizontal surface is W    mg x . Where m is mass of the body, x is the distance moved and  is coefficient of dynamic friction between the two surfaces in contact. Similarly, work done in moving a body up a rough inclined plane of inclination  is

W  mg  sin    cos   x 

When two vehicles of masses m1 , m2 moving with velocities v1 and v2 respectively are stopped by the

1 m1v12 x1 m v2 2   1 12 same force, the their stopping distances x1 and x2 are x 2 1 m2 v 2 m2 v 22 2 t1 m1v1 If t1 and t2 are times taken by the two bodies to stop, then t  m v 2 2 2 37