**Work **

Work is said to be done if a body gets displaced in the direction of applied force. It is a scalar quantity. Its unit is Joule (J) in S.I. and erg in C.G.S. It is also defined as the dot product of force ($\overrightarrow{F}$) and displacement ($\overrightarrow{S}$),

i.e. W = $\overrightarrow{F}$.$\overrightarrow{S}$

$\therefore $ W = FS Cos$\theta $ where $\theta $ is the angle between $\overrightarrow{F}$and $\overrightarrow{S}$.

**Special cases:**

1. When $\theta $ = 0º,

w = FScos$\theta $ = FScos0º = FS (+ve)

The positive sign shows that the work is done in the direction of applied force.

2. When $\theta $ = 180º

w = FScos180º = –FS (–ve)

The negative sign shows that the work is done in the direction opposite to applied force.

3. When $\theta $ = 90º

w = FScos90º = 0 (+ve)

The +ve sign shows that the work is done in the direction of applied force.

**Energy **

The capacity of doing work is known as energy. It’s unit is Joule (J) in S.I. and erg in C.G.S.

1J = 10^{7} erg

**Types of energy **

**1.**** ****Kinetic Energy (K.E.)**

It is the energy possessed by a body due to its motion. It is given by K.E. = $\frac{1}{2}$ mv^{2}.

**2.**** ****Potential Energy (P.E.)**

It is the energy possessed by a body due to its position. It is given by P.E. = mgh

**Mechanical energy **

It is the sum of kinetic energy (K.E.) and potential energy (P.E.)

**Power: **

The rate of doing work is called power. It is given by

P = $\frac{w}{t}$ = $\frac{\overrightarrow{F}.\overrightarrow{S}}{t}$ = $\overrightarrow{F}$. $\frac{\overrightarrow{S}}{t}$ =$\overrightarrow{F}$ . $\overrightarrow{V}$ = FVcos$\theta $

If the displacement is produced is the direction of applied force, i.e. $\theta $ = 0º,

then P = $\frac{w}{t}$ = FV

Its unit is Js^{–1} or watt (w)

1 K.W = 10^{3}w

1MW = 10^{6}w

1 H.P. = 746 w

1mW = 10^{–3}w

1$\mu $W = 10^{–6}w

1nw = 10^{–9}w

**Relation between K.E. and linear momentum (P)**

Let us consider a body of mass ‘m’ moving with velocity ‘v’. Then we have

Linear momentum (P) = mv

and K.E. = $\frac{1}{2}$ mv^{2}

= $\frac{1}{2}$ mv^{2$\times $$\frac{m}{m}$}

= $\frac{1}{2}$ $\frac{{{m}^{2}}{{v}^{2}}}{m}$

=$\frac{1}{2}$$\frac{{{(mv)}^{2}}}{m}$

$\therefore $ K.E. = $\frac{{{P}^{2}}}{2m}$

**Work – energy Theorem: **

Work-energy theorem states that the change in kinetic energy of a body is equal to work done.

Let us consider a body of mass ‘m’ moving with initial velocity ‘u’ at a point. If force ‘F’ is applied on it, it moves with constant acceleration ‘a’. After time ‘t’, its velocity becomes ‘v’ at point B. Let ‘S’ be the displacement covered by the body in time ‘t’.

Then, the amount of work done on moving from A to B is

W = Fs

Or, W = mas …………… (i) [F = ma]

We have,

v^{2} = u^{2} + 2as

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ …… (ii)

From (i) and (ii),

W = ma $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$

Or, W = $\frac{1}{2}$ mv^{2} –$\frac{1}{2}$ mu^{2}

Or, W = (K.E)_{final} – (K.E)_{initial}

$\therefore $W = $\Delta $K.E

which is the work energy theorem.

If the body is moving in vertical direction under the action of gravity, then

$\Delta $K.E. = $\Delta $P.E.

$\therefore $W = $\Delta $K.E. = $\Delta $P.E.

Formulae

1. W = FScos$\theta $

2. P = $\frac{W}{t}$ = FVcos$\theta $

For $\theta $= 0^{O }, P = Force× Velocity

3. K.E = $\frac{{{P}^{2}}}{2m}$

4. W = $\Delta $K.E = $\Delta $P.E

**Work done by Variable force **

A force is said to be variable if it changes with respect to position.

Let us consider a variable force F(x) is applied on a body and the body displaces from A to B in a fixed direction (X-axis). We can consider the entire displacement (AB) as a sum of a number of infinitesimal displacements. Let PQ = dx is one of the infinitesimal displacements. Then the small amount of work done in moving the body from P to Q is given by

dw = F × dx = PS × PQ

or, dw = Area of rectangle PQTS

For dx$\to $ 0, ST $\approx $ $\overset\frown{SR}$

∴ dw = area of strip PQRS ………… (i)

Now, the total work done in moving the object from A to B can be obtained by integrating equation (i) from x = x_{1} to x = x_{2}, we get,

w = $\int\limits_{x1}^{x2}{dw}$ = $\int\limits_{x1}^{x2}{area\text{ }of\text{ }strip\text{ }PQRS}$

$\therefore $ w = area of ABCDA

This shows that the work done by the variable force is numerically equal to the area between force curve and displacement axis.

**Work done by constant force **

Force is said to be constant if it does not change with respect to position.

Let us consider a body of mass ‘m’ moving with initial velocity ‘u’ at a point. If force ‘F’ is applied on it, it moves with constant acceleration ‘a’. After time ‘t’, its velocity becomes ‘v’ at point B. Let ‘S’ be the displacement covered by the body in time ‘t’.

Then, the amount of work done on moving from A to B is

W = Fs

Or, W = ma s …………… (i) [$\because $F = ma]

We have,

v^{2} = u^{2} + 2as

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ ………… (ii)

From (i) and (ii),

W = ma $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$

Or, W = $\frac{1}{2}$ mv^{2} –$\frac{1}{2}$ mu^{2}

Or, W = (K.E)_{final} – (K.E)_{initial}

$\therefore $W = $\Delta $K.E

Hence, Work done by constant force is equal to change in kinetic energy of the body.

**Principle of Conservation of energy (Mechanical):**

Principle of Conservation of energy states that energy can neither be created nor destroyed, but can be changed from one form to another form.

For a freely falling body, the total mechanical energy of the body remains constant throughout the motion.

Let us consider an object of mass ‘m’ is initially at rest at point A, which is at a height of ‘h’ from the ground level. The object is allowed to fall freely under the action of gravity. After covering a distance ‘x’, it reaches point B and finally reaches the ground level at point C.

**At position A**

Height = h and velocity = V_{A} = 0

$\therefore $ K.E_{A} = $\frac{1}{2}$ mV_{A}^{2} = 0

And P.E_{A} = mgh

Now, mechanical energy at point A = (M.E.)_{A} = K.E_{A} + P.E_{A}

= 0 + mgh

= mgh

$\therefore $ M.E_{A} = mgh ……….. (i)

**At position B**

Height = h – x and velocity = V_{B}

We know, V_{B}^{2} = V_{A}^{2} + 2aS

or, V_{B}^{2} = V_{A}^{2} + 2ghx

= 0 + 2gx

Now, K.E_{B }= $\frac{1}{2}$m (V_{B})^{2}

= $\frac{1}{2}$m × 2gx = mgx

P.E_{B} = mg(h – x)

Now, mechanical energy at B, (M.E)_{B} = K.E_{B} + P.E_{B}

= mgx + mg(h – x)

= mgx + mgh – mgx

= mgh

$\therefore $ M.E_{B} = mgh ………… (ii)

**At position C **

Height = 0 and velocity = V_{C}

We know,

V_{C}^{2} = V_{A}^{2} + 2gS

or, V_{C}^{2} = 0 + 2gh

Now, K.E_{C} = $\frac{1}{2}$ m(V_{C})^{2} = $\frac{1}{2}$m × 2gh = mgh

P.E_{C} = mgh × 0 = 0

Now, mechanical energy at C, M.E_{C} = K.E_{C} + P.E_{C}

= mgh

$\therefore $ M.E_{C} = mgh ……… (iii)

From (i), (ii) and (iii)

M.E_{A} = M.E_{B} = M.E_{C}

This shows that for a freely falling body, the total energy (mechanical) remains constant throughout the motion.

*Fig: Variation of K.E. and P.E. with height*

**One dimensional collision **

**a.**** ****Elastic collision **

A collision is said to be elastic if both linear momentum (P) and kinetic energy (K.E.) remain conserved. There is no loss of energy in elastic collisions.

Let us consider two objects of masses ‘m_{1}‘ and ‘m_{2}‘ moving with velocities ‘u_{1}‘ and ‘u_{2}‘ respectively. Let ‘u_{1}‘ is greater than ‘u_{2}‘, then the two objects collide and more separately with final velocities ‘v_{1}‘ and ‘v_{2}‘ respectively. Since the collision in elastic, from conservation of linear momentum.

m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

or, m_{1}(u_{1} – v_{1}) = m_{2}(v_{2} – u_{2}) ……….. (i)

And from conservation of kinetic energy,

$\frac{1}{2}$m_{1}u_{1}^{2} +$\frac{1}{2}$ m_{2}u_{2}^{2} =$\frac{1}{2}$ m_{1}v_{1}^{2} +$\frac{1}{2}$ m_{2}v_{2}^{2}

or, m_{1}(u_{1}^{2} – v_{1}^{2}) = m_{2}(u_{2}^{2} – v_{2}^{2}) ………. (ii)

Dividing (ii) by (i),

u_{1} + v_{1} = u_{2} + v_{2}

or, u_{1} – u_{2} = v_{2} – v_{1}

$\therefore $ u_{1} – u_{2} = v_{2} – v_{1}

Above result shows that velocity of approach is equal to the velocity of separation in elastic collision, i.e. relative velocity of before collision is equal to that after collision.

Further,

v_{1} = v_{2} + u_{2} – u_{1} ………. (iii)

And, v_{2} = v_{1} + u_{1} – u_{2} ……….. (iv)

From equation (i) and (iii),

m_{1}[u_{1} – (v_{2} + u_{2} – u_{1})] = m_{2}(v_{2 }– u_{2})

or, m_{1}[u_{1} – v_{2} – u_{2} + u_{1}) = m_{2}v_{2} – m_{2}u_{2}

or, 2m_{1}u_{1} – m_{1}u_{2} – m_{1}v_{2} = m_{2}v_{2} – m_{2}u_{2 }

or, 2m_{1}u_{1} – m_{1}u_{2} + m_{2}u_{2} = m_{1}v_{2} – m_{2}v_{2 }

or, 2m_{1}u_{1} – u_{2}(m_{1} – m_{2}) = v_{2}(m_{1} + m_{2})

or, v_{2 }= $\frac{2{{m}_{1}}u{}_{1}+{{u}_{2}}({{m}_{2}}-{{m}_{1}})}{{{m}_{1}}+{{m}_{2}}}$……….. (v)

From (i) and (iv),

m_{1}(u_{1 }– v_{1}) = m_{2}[(v_{1} + u_{1} – u_{2}) – u_{2}]

or, m_{1}u_{1} – m_{1}v_{1} = m_{2}v_{1} + m_{2}u_{1} – 2m_{2}u_{2}

or, m_{21} – m_{2}u_{1} + 2m_{2}u_{2} = m_{2}v_{1} + m_{1}v_{1}

or, 2m_{2}u_{2} + u_{1} (m_{1} – m_{2}) = v_{1}(m_{1} + m_{2})

or, v_{1} =$\frac{2{{m}_{2}}u{}_{2}+{{u}_{1}}({{m}_{1}}-{{m}_{2}})}{{{m}_{1}}+{{m}_{2}}}$ ……….. (vi)

From (v) and (vi) give the final velocities of m_{2 }and m_{1 }respectively.

*Special cases *

**1.**** ****Case I **

When m_{1} = m_{2} = m (Say).

Then from equations (v) and (vi), we get,

v_{1} = $\frac{2m{{u}_{2}}}{2m}$

$\therefore $ v_{1 }= u_{2}……….. (vii)

And, v_{2} = $\frac{2m{{u}_{1}}}{2m}$

$\therefore $ v_{2 }= u_{1}……….. (viii)

Equations (vii) and (viii) show that in elastic collisions the colliding bodies exchange their velocities if they have the same masses.

**2.**** ****Case II **

When m_{1} >>>> m_{2} and u_{2} = 0

Then from equations (v) and (vi), we get,

v_{1} $\approx $ $\frac{0+{{u}_{1}}{{m}_{1}}}{{{m}_{1}}}$

$\therefore $v_{1$\approx $} u_{1}

And, v_{2} $\approx $ 2u_{1}

**3.**** ****Case III **

When m_{2} >>>> m_{1} and u_{2} = 0

Then from equations (v) and (vi), we get,

v_{1} $\approx $ – u_{1}

And v_{2} $\approx $ 0

*Q.1.** **Show that in one dimensional elastic collision, velocity of approach is equal to velocity of separation. *

*i.e. [u*_{1}* – u*_{2}* = v*_{2 }*– v*_{1}*]*

*Q.2.** **Show that in one dimensional elastic collision, the colliding bodies exchange their velocities (if they have the same masses). i.e. [u*_{1}* = v*_{2}* and u*_{2}* = v*_{1}*]*

**b ****Inelastic collision **

The collision is said to be inelastic if linear momentum is conserved but not the kinetic energy. In inelastic collision, there is loss of kinetic energy.

Let us consider an object of mass ‘m_{1}‘ moving with initial velocity ‘u_{1}‘ collides with another object of mass ‘m_{2}‘, which is initially at rest (i.e. u_{2} = 0). After collision, the colliding bodies combine and move together with common velocity ‘v’. Since the collision is inelastic, only linear momentum is conserved, not the kinetic energy.

From the conservation of linear momentum,

m_{1}u_{1} + m_{2}u_{2} = m_{1}v + m_{2}v

Or, m_{1}u_{1} + 0 = (m_{1 }+ m_{2})v [$\because $ m_{2} is initially at rest]

$\therefore $ v = $\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}$……… (i)

The sum of K.E. before collision is

K.E._{1} = $\frac{1}{2}$m_{1}u_{1}^{2} + $\frac{1}{2}$m_{1}u_{2}^{2}

$\therefore $ K.E._{1} = $\frac{1}{2}$m_{1}u_{1}^{2 } …. (ii)

Now, the sum of K.E. after collision is

K.E._{2} = $\frac{1}{2}$m_{1}v^{2} + $\frac{1}{2}$m_{2}v^{2}

$\therefore $ K.E._{2} = $\frac{1}{2}$(m_{1} + m_{2})v^{2}…… (iii)

Dividing (ii) by (iii)

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}}){{v}^{2}}}{{{m}_{1}}{{u}_{1}}^{2}}$

Using equation (i),

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}})}{{{m}_{1}}{{u}_{1}}^{2}}{{(\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}})}^{2}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ < 1

$\therefore $K.E._{2} < K.E._{1}

Above relation shows that the kinetic energy after collision is less than that before energy, i.e. there is loss in kinetic energy in inelastic collision.

**Conservative Force **

A force is said to be conservative if work done by it in a closed path (loop) is zero. The work done by conservative force is path independent.

e.g. gravitational force, electrostatic force, magnetic force.

**Non–conservative Force **

A force is said to be non – conservative if work done by it in a closed path (loop) is not zero. The work done by non – conservative force is path dependent.

e.g. viscous force, frictional force, etc.

**Coefficient of Restitution (e):**

It is defined as the ratio of velocity of separation to the velocity of approaches. It is denoted by ‘e’ and given by

e = $\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}$

(i) For perfectly elastic collision, e = 1

(ii) For perfectly inelastic collision, e = 0

(iii) Practice in collision, 0 < e < 1

**Work, Energy & Power**

**Numerical Problems**

Q. No: 1, 2, 3, 4, 5, 6 are similar

Q.1. A stationary mass explodes into two parts of mass 4 units and 40 units respectively. If the larger mass has an initial K.E. 10 J, what is the initial kinetic energy of the smaller mass? Ans: 100 J

Q.2. A stationary mass explodes into two parts of mass 4 kg and 40 kg. The initial kinetic energy of larger mass is 10 J. Find the velocity of the smaller mass. Ans: 7.07 m/sec

Q.3. A stationary mass explodes into two parts of mass 4 units and 40 units respectively, If the larger mass has an initial K.E. 100J, what is the initial K.E. of the smaller mass? Ans: 1000 J

Q.4. An explosive of mass M placed at a point explodes into one-third and two-third parts. If the initial kinetic energy of the smaller part is 1000 J. What will be the initial K.E. of the bigger part? Ans: 500 J

Q.5. A ball of mass 4 kg moving with a velocity 10 ms^{-1} collides with another body of mass 16 kg moving with 4 ms^{-1} from the opposite direction and then coalesces into a single body. Compute the loss of energy on impact. Ans: 313.6J

Q.6. A bullet of mass 10 g is fired from a gun of mass 1 kg with a velocity of 100 ms^{-1}. Calculate the ratio of the kinetic energy of the bullet and the gun. Ans: 100 : 1

Q.7, 8 are similar

Q.7. A 0.15 kg glider is moving to the right a frictionless horizontal air track with a speed of 0.80 ms^{-1} .It has a head on collision with a 0.300 kg glider that is moving to the left with a speed of 2.2 ms^{–1}. Find the final velocity (magnitude and direction) of each glider if the collision is elastic. Ans: 3.2ms^{-1} and 0.2ms^{-1} (both towards left)

Q.8. A ball A of mass 0.1 kg moving with a velocity of 6ms^{-1} collides directly with a ball B of mass 0.2 kg at rest. Calculate their common velocity if both balls move off together. If ball A had rebounded with a velocity of 2ms^{-1} in the opposite direction after collision, what would be the new velocity of B? Ans: 2 m/s, 4 m/s

Q.9, 10, 11, 12 are similar

Q. 9 A typical car weighs about 1200 N. If the coefficient of rolling friction is m_{r} = 0.015. What horizontal force is needed to make the car move with constant speed of 72 km/hr on a level road? Also calculate the power developed by the engine to maintain this speed. Ans. 18 N, 360 W

Q.10. A car of mass 1000 kg moves at a constant speed of 20 m/s along a horizontal road. If a constant frictional force of 200 N. is acting between the car and the road. Calculate the power developed by the engine. Ans: 4 KW

Q. 11 A car of mass 1000 kg moves at a constant speed of 25 m/s along a horizontal road where frictional force is 200 N. Calculate the power developed by the engine. Ans: 5 kw

Q.12. A car of mass 1000 kg. moves at a constant speed of 20 ms^{-1} along a horizontal road where the friction force is 200 N. Calculate the power developed by the engine. Ans: 4 KW

Q.13, 14, 15 are similar

Q.13. A 650 KW power engine of a vehicle of mass 1.5 x 10^{5} Kg is rising on an inclined plane of inclination 1 in 100 with a constant speed of 60 km/hr. Find the frictional force between the wheels of the vehicle and the plane. Ans: 24000 N

Q.14. A train of mass 2 x 10^{5} kg moves at a constant speed of 72 kmh^{-1} up a straight inclined against a frictional force of 1.28 x 10^{4}N. The incline is such that the train rises vertically 1.0 m for every 100 m travelled along the incline. Calculate the necessary power developed by the train. Ans: 656 KW

Q.15. Find the power of an engine in kilowatts, which pulls a train of mass 600 tonnes up an incline of 1 in 100 at the rate of 60 km/hr. The weight of the engine is 200 tonnes and the resistance due to friction is 50 Newton’s per tonne. Ans: 2000 KW

Hint: 1 tonne = 1000 kg

Q.16. A water reservoir tank of capacity 250 m^{3} is situated at a height of 20 m from the water level. What will be the power of an electric motor to be used to fill the tank in 3 hours? Efficiency of motor is 70%. Ans: 6613.76 Watt

Q.17. A block of weight 150N is pulled 20 m along a horizontal surface at constant velocity. Calculate the work done by the pulling force if the coefficient of Kinetic friction is 0.20 and the pulling force makes an angle of 60° with the vertical. Ans: 537.93 J

Q.18. A bullet of mass 20 g. travelling horizontally at 100 ms^{-1} embeds itself in the centre of a block of wood mass 1 kg, which is suspended by light vertical string 1 m. in length. Calculate the maximum inclination of the string to the vertical. Ans: 35.9°

Q.19. You throw a 20 N rock vertically into the air from ground level. You observe that when it is 15 m above the ground, it is travelling at 25 m/s upward. Use the work-energy theorem to find (i) its speed as it left the ground and (ii) its maximum height. Ans: 30.41 m/sec, 46.24 m

Q.20. The constant force resisting the motion of a car of mass 1500 kg is equal to one fifteenth of its weight if, when travelling at 48 km/h, the car is brought to rest in a distance of 50 m by applying the brakes, find the additional retarding force due to the brakes (assumed constant) and heat developed in the brakes. Ans: 1666.67 N, 83333.5 J

Also Read: Laws of Motion Class 11