Work, Energy, Power |

 Work

Definition:

Work is said to be done on a body when an external force displaces a body in the direction of applied force.

S.I. Unit: Joule

1 joule = (1newton) (1 meter) or 1 J = 1 N . m

C.G.S unit = erg

1 J = $10^7$ erg

Dimension: $ML^2T^{-2}$

Mathematical Definition

Work (W) = $\overrightarrow{F} \cdot \overrightarrow{s}$

$\to$ Mathematically Work is the result of a dot product of two vectors i.e. force and displacement.

$\to$ Work is a scalar quantity.

Conditions

Work (W)= $FS \cos \theta$

$\to$ $\theta$ is the angle between force and displacement.

$\to$ $\theta= 0^{\circ}, W= FS$ displacement is in direction of force, the work done is maximum.

$\to$ $\theta= 90^{\circ} W=0$ displacement is perpendicular direction of force, the work done is zero.

$\textbf{For example}$

I . Work done by the centripetal force in displacing a particle alone a circular path is zero.

ii . Work done by centripetal force ( i.e gravitational pull ) in revolving satellite around the earth is zero.

iii. When a person carrying a load on the head moves over a horizontal work done against the gravitational force is zero.

iv. When a car moves with a uniform speed over a frictionless road, work done is zero.

$\to 180^{\circ} >\theta>90^{\circ} W= -ve$ work done is negative.

$\to$ Work done by friction and when the body is thrown up the work by gravitational pull is negative.

Work Done by Variable Force

If the force is the function of displacement, then, the work done can be found using integration

$W = \displaystyle{\int_{x=a}^{x=b}} F_xdx$

Energy

Definition:

The capacity of a body to do work is called energy.

S.I. Unit: Joule

C.G.S Unit: erg

Dimension: $ML^2T^{-2}$

Mechanical Energy

The energy possessed by the body due to its motion or position is called mechanical energy. They are kinetic energy and mechanical energy.

Kinetic Energy

The energy possesed by the body due to its motion is called kinetic energy.

$E = \dfrac{1}{2} mv^2 = \dfrac{p^2}{2m}$

$m$ is the mass of the body, $v$ is the velocity of the body and $p$ is the momentum of the body.

Hence For constant momentum, K.E is inversely proportional to the mass of the body.

Potential Energy

The energy possessed by a body due to its position or configuration is called potential energy.

$\to$ Gravitational Potential Energy

For the height h above the surface of the earth: $P.E. = - \dfrac{GMm}{R+h} = - \dfrac{mgR^2}{R+h}$, Where M = Mass of the earth, r = Radius of the earth, h= height above the earth

The change in potential energy when a body is taken from surface to height h is $\Delta PE = -\dfrac{mgR^2}{R+h} + \dfrac{mgR^2}{R} = \dfrac{mgRh}{R+h}$

If $h<< R$ $\Delta PE = mgh$

If $h>> R$ $\Delta PE = mgR$

$\to$ Elastic Potential Energy

The energy stored in the spring when it is stretched by a force (F) producing an extension of (x) is given as

$U = \dfrac{1}{2} Fx = \dfrac{1}{2} kx^2$

$\to$ KE is never negative but PE can be positive, negative and zero.

Work-Energy Theorem

$\to$ The work and energy are related and are equivalent.

$\to$ Work = Change in KE = $\dfrac{1}{2} mv^2 - \dfrac{1}{2} mu^2$

Conservation of energy

$\to$ The work-energy theorem is based on the conservation of energy.

$\to$ It states that energy can neither be created nor be destroyed but can be transferred from one form to another.

K.E. + P.E. = Constant

Work Done by Conservative and Non-Conservative Forces

$\to$ Conservative Forces

The work done by a conservative force is independent of path.

It depends on the final and initial state.

The work done in the closed path is zero.

Total mechanical energy is conserved

eg. Gravitational force, elastic force etc.

$\to$ Conservative Forces

The work done by a nonconservative force is dependent on the path taken.

The work done in the closed path is not zero.

Total mechanical energy is not conserved

eg. frictional force, viscous force etc.

Work and Energy

$\to$ Whenever work is done by a body, the work is + ve and energy decreases.

$\to$ Whenever work is done on a body, work is - ve, and its energy increases.

Power

Definition

The work done per unit time is known as power.

Power = $\dfrac{\text{Work Done}}{\text{Time Taken}}$

It is a scalar quantity.

S.I. Unit is Watt (W)

CGS Unit is ergs/s

Practical Unit: Horse Power (HP), 1 HP = 746 Watt

Dimension: $ML^2T^{-3}$

1 hp = 746 W = 0.746 kW hp= Horse power

Mathematical Definition

Work (W) = $\overrightarrow{F} \cdot \overrightarrow{v} = Fv \cos \theta$

$\to$ Mathematically Power is the result of a dot product of two vectors i.e. force and velocity.

$\to$ Power is a scalar quantity.

$\to$ It is also defined as rate of change of energy $\dfrac{\Delta E}{t}$

Collision

The interaction between two or more bodies for a short time after which their kinetic energy and momentum are changed is called collision.

Elastic Collision:

$\to$ Kinetic energy and linear momentum are conserved.

$\to$ The force creating elastic collision are conservative in nature.

$\to$ Mechanical Energy is conserved.

Inelastic Collision

A collision in which the total kinetic energy after the collision is less than before the collision is called an inelastic collision i.e. K.E is not conserved.

Linear momentum is conserved.

• Are there any examples of perfectly inelastic collisions?

$\to$The ballistic pendulum is a practical device in Qwhich an inelastic collision takes place. Until the advent of modern instrumentation, the ballistic pendulum was widely used to measure the speed of projectiles.

(Two bodies stick after perfect inelastic collision)

Perfectly inelastic collision: If the entire K.E. of bodies is converted to another form of energy after collision.

Laws of Collision

The velocity of separation between particles after the collision is directly proportional to the velocity of approach of these particles before the collision.

$u_1-u_2$ = velocity of approach

$v_2 – v_1$ = velocity of separation

According to the law of collision

$v_2 - v_1 \propto ( u_1 - u_2 )$

$v_2 - v_1 = e ( u_1 - u_2 )$

Where e is proportionality constant depending on the nature of the collision and is called the coefficient of restitution.

It is unitless and dimensionless.

Coefficient of restitution

The ratio of the velocity of separation after a collision to the velocity of approach before the collision is known as the coefficient of restitution.

Coefficient of restitution ( e ) = $\dfrac{\text{velocity of separation}}{\text{velocity of approach}} = \dfrac{(v_2 - v_1 )}{(u_1 - u_2 )}$

$\to$ For elastic collision, $e = 1$

$\to$ For perfectly inelastic collision , $e = 0$

$\to$ For collision in practices , $0 < e < 1$

[ Note : Linear momentum is conservative in every collision . Total energy is conservative in every collision . Kinetic energy is conservative only in elastic collision . ]

Elastic collision in one dimension

$\textbf{Conservation of momentum} m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ....(i)

$\textbf{Conservation of Energy} \dfrac{1}{2} m_1u_1^2 + \dfrac{1}{2} m_2u_2^2 = \dfrac{1}{2} m_1v_1^2 + \dfrac{1}{2} m_2v_2^2$ ... (ii)

$v_2 - v_1 = u_1 - u_2$ i.e. velocity of separation = velocity approach, $e = 1$

It implies that particles moving slowly before collision becomes fast after the elastic collision and vice-versa.

Solving equations (i) and (ii)

$\to v_1 = \dfrac{(m_1 -m_2 )}{(m_1 + m_2)} u_1 + \dfrac{2m_2}{(m1 + m2 )} u_2$

$\to v_2 = \dfrac{ 2m_1}{(m_1 + m_2) }u_1 - \dfrac{(m_1 - m_2)}{(m_1 + m_2) }u_2$

Special cases of elastic collision

A. When particle B is at rest ( i.e. $u_2 = 0$ )

Then

$v_1 = \dfrac{(m_1 -m_2 )}{(m_1 + m_2)} u_1$ and

$\to v_2 = \dfrac{ 2m_1}{(m_1 + m_2) }u_1$

B . When their masses be equal ( i.e . $m_1 = m_2 = m$)

Then $v_2 = u_1$ and $v_1 = u_2$ i.e, exchange their velocities .

C. When $m_1 = m_2$ and particle B is Stationary i.e. $u_2 = 0$ then $v_1 = 0$ and $v_2 = u_1$

D . When panicle A is very tiny and B is at rest i.e.

$m_1 <<< m_2$ and $u_2 = 0$, then $v_1 = - u_1$ and $v_2 = 0$

E. When body A is huge mass ($m_1 >> m_2$ ) , and B is at rest then $v_1 = u_1$ and $v_2 = u_2$.

Perfectly inelastic collision in one dimension :

When two bodies of masses $m_1$ and $m_2$ moving at velocities $u_1$ and $u_2$ get perfect inelastic collision and move with common velocity v ,then

A. $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

$v = \dfrac{m_1 u_1 + m_2 u_2 }{m_1 +m_2}$

B. Coefficient of restitution =0.

C. Loss in Kinetic Energy = $\dfrac{1}{2} \dfrac{m_1m_2}{m_1 + m_2} (u_1 -u_2)^2$

Inelastic collision of a body dropped from a height

When a body is dropped from a height $h_0$, gets inelastic collision with earth surface and the body rebounds to height $h_1$ then

A. Striking velocity of the body just before collision $u_1= u = \sqrt{2gh_0}$ ( downward )

B. Velocity of bounce $v_1=v = \sqrt{2gh_1}$ ( upward )

c. Velocity of earth after and before collision be each zero i.e. $u_2 = 0 = v_2$.

D. Coefficient of restitution $e = \dfrac{(v_2 - v_1 )}{(u_1 - u_2 )} = \dfrac{(0 - ( - v ))}{(u - 0)} = \dfrac{v}{u} = \sqrt{\dfrac{h_1}{h_0}}$

E. Height raised after bounce $h_1 = e^2 h_0$

F. Velocity after bounce $v = ue$.

G. K.E. after bounce $E = \dfrac{1}{2} mv^2 = \dfrac{1}{2} mu^2e^2 = e^2E_0$

H . Fractional loss in momentum: $\dfrac{∆p}{p_0} = \dfrac{(p_0 - P)}{p_0} = 1 - e$

I. Fractional loss in K.E. $\dfrac{∆ E}{ E_0} = \dfrac{(E_0 -E )}{E_0} = 1 - e^2$

When a body released from a height $h_0$ , bounces after successive collisions and finally comes to rest after an infinite number of collisions provided that e be the coefficient of restitution for each collision then

A . Height raised after $n^{th}$ bounces: $h_n = ( e^{2n} ) h_0$

B. Displacement covered till rest is $h_0$.

C. Distance covered till rest: $s = \dfrac{(1+e^2)}{(1-e^2)} h_0$

D. Time taken till rest: $T = \dfrac{(1 + e)}{(1 - e)} \sqrt{\dfrac{2h_0}{g}}$

E. Fractional loss in velocity $∆v/v = 1 - e^n$ and

F. Fractional loss in momentum

$∆p/p = 1 – e^n$

G. K.E. lost after the nth bounce

$∆ E = ( 1 – e^{2n} ) E_0$

Better You Know

When a particle ( mass m and velocity v ) strikes normally with a wall and returns with the same velocity, then the change in momentum is 2mv.

A bullet of mass M hits a block of mass M’ . The transfer of energy is maximum when M’ = M.

When a man rowing a boat upstream 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.

A bullet is fired from a rifle . If rifle recoils freely , the K.E of the rifle is less than that of bullet .

When an air bubble rises to the surface of the water from below. PE decreases.

Only momentum not kinetic energy is conserved in inelastic collision while both momentum and K.E are conserved in elastic collision .

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

If a shell fired for a cannon explodes in mid-air then its total kinetic energy increases but total momentum is conserved .

When two like charges are brought together, P.E increases ( proton-proton, electron-electron ), and when unlike charges are brought together, P.E decreases ( proton-electron ).

Weight of hydrogen balloon in air is negative . Hence gravitational P.E of it decreases with altitude .

The force required to hold the machine gun in position $F = np = m \sqrt{2m K}$ where m is the mass of each bullet

When a machine gun fires n bullet per second, each with K.E. ‘K’ then the power of the machine gun is P =nK

Work is necessarily done if energy of the body gets changed .

If E = 0 then p = 0 i.e. a body cannot have linear momentum without K.E and vice-versa

When a body is in static or dynamic equilibrium , then work done is zero .

P.E. decrease on the rise of an air bubble in water because work is done by upthrust.

The potential energy of a system in the state of unstable equilibrium is maximum .

If a man pushes an object but fails to displace it, he does no work at all.

When a body moves with a constant speed around a circle , then no work is done on it .

Work done by the centripetal force is always zero.

The velocity of a particle when its momentum is equal to its K.E is 2 m/s .

i. When $p_1 = p_2$, = $\sqrt{2m_1E_1}= \sqrt{2m2E_2}$

i.e. K.E of lighter body will be more .

ii. When $E_1 = E_2$

$p_1^2 / 2m_1 = p_2^2 / 2m_2$

$or \dfrac{p_1}{p_2} = \sqrt{\dfrac{m_1}{m_2}}$

i.e. when kinetic energies of two bodies are the same, then the momentum of the heavier body will be more.

iii. K.E depends on the frame of reference.

For example, K.E of a person of mass m sitting in a train moving with a speed v is zero in the frame of the train but $\dfrac{1}{2} mv^2$ in the frame of earth.

The slope of work - time curve at any instant gives us the power at that instant , as P = dW / dt = tan θ Work done by gravity is + ve if body moves downwards and is -ve if body moves upwards .

A body is moved along a straight line by a machine delivering constant power then the relationship between displacement , velocity and acceleration with time may be obtained as follows :

$P = Fv$

or, $P = mav$

or. $P = m v/t v$

or. $\dfrac{v^2}{t}$ = constant

or. $v^2 \propto t$

$v \propto t^{1/2}$

$s \propto t^{3/2}$

$a \propto t^{-1/2}$

When a chain of length ‘l’ lies on a table with l/n th part hanging , then work done to pull the hanging part on the table

$W = \dfrac{mgl}{2n^2}$

A uniform iron chain of mass m and length I held on a smooth table with l/n part overhung is released. The end of the chain leaves the edge after falling under gravity with velocity.

$v = \dfrac{1}{n} \sqrt{(n^2-1)gl}$

Change in gravitational P.E of a body of mass m at a height h above the surface of earth is

P.E = $mgh [ \dfrac{R}{(R + h)} ] = mgh$ , when $h <<< R$.

When water is flowing through a pipe with a speed v, then the rate at which K.E. is being imparted to the liquid is

K.E. = $\dfrac{1}{2} \rho A v^3 \propto v^3$

Power $P \propto v^3$

[ Hint: Volume flowing/sec = Av

Mass flowing / sec , $\dot{m} = Av \rho$

K.E. / sec $= \dfrac{1}{2} \dot{m} v^2 =\dfrac{1}{2} \rho A v^3$

A particle of mass m moves from rest under the action of a constant force F for t seconds

The maximum power attained is $\dfrac{(F^2 t)}{m}$

Work done to empty a tank filled with water to half depth is

W = 3/8 Mgh

[ Hint : $Mg (h/2) - M/2 g ( h/4 ) = 3 Mgh/8$]

Work done to take all the liquid out of the full tank, W = Mgh/2

A rod of mass m and length is lying on a horizontal table. The work done in making it stand on one end will be W = mgl/2 ( since C . G . is raised by l/2 )

A stick of length l pivoted at one end is displaced through an angle θ. Then increase in P.E is

∆U = mgl / 2 ( 1 - cos θ )

When a pendulum bob of mass m is displaced through an angle θ from the mean position. Then increase in P.E is

∆ U = mgl ( 1 – cos θ )

When two vehicles of masses $m_1$ and $m_2$ moving with velocities $v_1$ and $v_2$ respectively are stopped by the same force , then stopping distances $s_1$ and $s_2$ are

$\dfrac{s_1}{s_2} = \dfrac{1/2 m_1v_1^2)}{(1/2 m_2v_2^2 )} = \dfrac{(m_1v_1^2)}{(m_1v_2^2 )}$

If $t_1$ and $t_2$ are times taken by the two bodies to stop, then $\dfrac{t_1 }{t_2} = \dfrac{m_1v_1}{ m_2v_2}$ .

If a car of mass $m_c$ and a bus of mass $m_b$ having kinetic energies $K_c$ and $K_b$ are stopped, under the action of same retarding force in distances $S_c$ and $S_b$ in times $t_c$ and $t_b$ respectively then:

i. If $K_c = K_b$;

$S_c = S_b$ and $\dfrac{t_c }{ t_b} = \sqrt{\dfrac{m_c}{m_b}}$

i.e. bus takes more time to stop, but they cover the same distance before stopping

ii. If $p_c = p_b$, then $\dfrac{S_c}{S_b} = \dfrac{m_b}{m_c}$ and $t_c = t_b$

i.e. stopping distance for car is more than that for the bus but stopping time is the same

A car of mass $m_1$ and a bus of mass $m_2$ moving with the same velocity ( v ). If they are brought to rest by the application of brakes, which provide equal retardation, then both will stop at the same distance.

In the above statement, if the road were rough, stopping distance is given by

$s = \dfrac{v^2 }{2μg}$

Stopping time

$t = \dfrac{v}{\mu g}$

Two springs having force, constants $k_1$ and $k_2$ are extended through the same distance x. Then the ratio of their elastic energies will $k_1: k_2$

A shell initially at rest explodes into two pieces of equal mass, the two pieces will move with the same velocity in opposite directions.

[ Hint : $MV = mv_1 + mv_2$

or. $0 = mv_1 + mv_2$

$v_1 = -v_2$

A body of mass $m_1$ moving with velocity $u_1$ strikes head-on ( elastically ) with another body of mass $m_2$ initially at rest. Then the fraction of energy lost ( or transferred ) by the moving body in the collision itself is $\dfrac{(4 m_1m_2 )}{(m_1 + m_2)^2}$

A body of mass $m_1$ moving with velocity $u_1$ strikes another body of mass $m_2$ initially at rest, inelastically. Then the fraction of energy lost by the moving body in the collision itself is $\dfrac{m_2}{(m_1 + m_2)}$ .

If a stationary bomb explodes into two masses $m_1$ and $m_2$ with kinetic energies $E_1$ and $E_2$ then,

$E_1 = \dfrac{m_2}{(m_1 + m_2)} \times$ total energy

$E_2 = \dfrac{m_1}{(m_1 + m_2)} \times$ total energy

For two springs A and B , if $K_A > K_B$

i. $W_A > W_B$ when they are stretched by the same amount.

ii . $W_A < W_B$ when they are stretched by the same force.

Elastic potential energy ( U ) =$\dfrac{1}{2} kx^2$ where ( k = force constant of spring )

If a spring is already stretched by amount $x_1$, and is further stretched by amount $x_2$, then work done in stretching the spring from $x_1$ to $x_2$ is

$W = \dfrac{1}{2} k (x_1 + x_2)^2 -\dfrac{1}{2} kx_1^2$

A bullet loses 1/n th of its kinetic energy after penetrating a plank. Then the number of planks required to stop the bullet is n.

A bullet loses 1/n th of its velocity after penetrating x distance of a plank . Then

i. Further distance travelled by the bullet before coming to rest $s_1 = \dfrac{( n-1 )^2 x}{(2n -1)}$

ii. Total distance travelled by the bullet before coming to rest $s = \dfrac{(n^2 x)}{(2n-1 )}$.

A bullet loses 1/n th of its kinetic energy after penetrating x distance of the plank. Then

i. Further distance traveled by the bullet before coming to rest, $s_1 = (n – 1 ) x$

ii. Total distance traveled by the bullet before coming to rest $s = nx.$

A bullet loses 1/n th of its velocity after penetrating a plank . Then the number of planks required to stop the bullet will be

i. n/2 +1 ( if n is even )

ii. n + 0.5 ( if n is odd )