Force \(\overrightarrow { F } =mg\left( -\hat { j } \right) =-mg\hat { j } \)
Displacement vector \(d\vec{r}\) = dx\(\hat { i } \) + dy \(\hat { j } \)
(As the displacement is in two dimension; unit vectors \(\hat { j } \) and \(\hat { i } \)  are used)
(a) Since the motion is only vertical, horizontal displacement component dx s zero. Hence, work done by the force along path 1 (of distance h).
\({ W }_{ push\ \ 1 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dy\hat{j})=-mg\int _{ 0 }^{ h }{ dy=-mgh } } } \)
Total work done for path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ C }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } } } } } \)
But \(\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
\(\int _{ A }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ C }^{ D }{ (-mg\hat { j } ).(dy\hat { j } )=-mg\int _{ 0 }^{ h }{ dy } =-mgh } } \)
\(\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
Therefore, the total work done by the force along the path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \vec { F } } .d\overrightarrow { r } =-mgh\)
Note that the work done by the conservative force is independent of the path.

Given:
F = 50N
m₁ = 5 kg
m₂ = 10 kg
m₃ = 15 kg
All the blocks move with common acceleration a under the force F = 50N
∴ F = (m₁ + m₂ + m₃)a
or a = \(\frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } =\frac { 50 }{ 5+10+15 } =\frac { 5 }{ 3 } \) ms^-2
To determine T₁ Refer to the free-body diagram for m₁ shown in the diagram. Clearly, the tension T₁ produces acceleration a in mass m₁.
∴ T₁ = m₁a = \(5\times \frac { 5 }{ 3 } =\frac { 25 }{ 3 } \) = 8.33 N

To determine T₂ Refer to the free-body diagram for m₃ shown in the diagram (b). Force F acts towards right and tension T₂ acts towards left.
∴ F - T₂ = m₃ a (or) 50 - T₂ = 15 x \(\frac { 5 }{ 3 } \) or T₂ =25N

When a vehicle travels in a curved path, there must be a centripetal force acting on it. This centripetal force is provided by the frictional force between tyre and surface of the road. Consider a vehicle of mass 'm' moving at a speed 'v' in the circular track of radius 'r'. There are three forces acting on the vehicle when it moves as shown in the figure.
(i) Gravitational force (mg) acting downwards
(ii) Normal force (mg) acting upwards
(iii) Frictional force (F_s) acting horizontally inwards along the road
Suppose the road is horizontal then the normal force and gravitational force are exactly equal and opposite. The centripetal force is provided by the force of static friction Fs between the tyre and surface of the road which acts towards the center of the circular track,
\(\frac { m{ v }^{ 2 } }{ r } ={ F }_{ s }\)
As we have already seen in the previous section, the static friction can increase from zero to a maximum value
F_s ≤ μ_smg
There are two conditions possible namely without skidding and skidding.
For without skidding \(\frac { m{ v }^{ 2 } }{ r } \le { \mu }_{ s }mg\), or \({ \mu }_{ s }\ge \frac { { v }^{ 2 } }{ rg } \) or \(\sqrt { { \mu }_{ s }rg } \ge v\)
The static friction would be able to provide necessary-centripetal force to bend the car on the road.

Consider two blocks of masses m₁ and m₂ (m₁  > m₂) kept in contact with each other on a smooth, horizontal frictionless surface as shown in the figure.

By the application of a horizontal force F, both the blocks are set into motion with acceleration 'a' simultaneously in the direction ofthe force F.
To find the acceleration \(\vec { a } \), Newton's second law has to be applied to the system (combined mass m = m₁ + m₂)
\(\vec { F } =m\vec { a } \)
If we choose the motion of the two masses along the positive x direction
\(F\hat { i } =ma\hat { i } \)
By comparing components on both sides of the above equation
F = ma where m = m₁ + m₂
The acceleration of the system is given by
∴ a = \(\frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } \) ..............(1)
The force exerted by the block m₁ on m₂ due to its motion is called force of contact (\(\vec { f_{ 21 } } \)). According to Newton's third law, the block m₂ will exert an equivalent opposite reaction force (\(\vec { f_{ 12 } } \)) on block m₁..Figure shows the free body diagram of block m₁.
Figure shows the free body diagram of block m₁
∴ \(F\hat { i } ={ \vec { f } }_{ 12 }\hat { i } ={ m }_{ 1 }a\hat { i } \)
By comparing the components on both sides. of the above equation, we get
F-f₁₂ = m₁a
f₁₂ = F-m₁a      .....(2)
Substituting the value of accelera~ion from equation (1) in (2) we get
f₁₂= \(F-{ m }_{ 1 }\left( \frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
f₁₂=\(f\left[ 1-\frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right] \)
f₁₂=\(\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)    ....(3)
Equation (3) shows that the magnitude of contact force depends on mass m: which provides the reaction force. Note that this force is acting along the negative x direction.
In vector notation, the reaction force on mass m₁ is given by \(\vec { { fi }_{ 12 } } =\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
For mass m₂ there is only one force acting on it in the x direction and it is denoted by f₂₁ This force is exerted by mass m₁. The free body diagram for mass m₂ is shown in the figure.

Applying Newton's second law for mass m₂
\(f_{ 21 }\hat { i } ={ m }_{ 2 }a\hat { i } \)
By comparing the components on, both sides of the above equation
f₂₁ = m₂a     .....(4)
Substituting for acceleration from equation (1) in equation (4), we get \({ f }_{ 21 }=\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
In this case the magnitude of the contact force is \({ f }_{ 21 }=\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \).
Free body diagram of block of mass m₂

When an object of mass m slides on a frictionless surface inclined at an angle e as shown in the figure, the forces acting on it decides the
(a) acceleration of the object 
(b) speed of the object when it reaches the bottom.
The force acting on the object is
(i) Downward gravitational force (mg)
(ii) Normal force perpendicular to inclined surface (N)

To draw the free body diagram, the block is assumed to be a point mass [in figure (a)]. Since the motion is on the inclined surface, we have to choose the coordinate system parallel to the inclined surface as shown
in the figure (b).
The gravitational force mg is resolved in to parallel component mg sine along the inclined plane and perpendicular component mg coss perpendicular to the inclined surface [figure (b)].
Note that the angle made by the gravitational force (mg) with the perpendicular to the surface is equal to the angle of inclination e' as shown in figure (c).
There is no motion (acceleration) along the y axis. Applying Newton's second law in the y direction
\(-mg\ cos\theta \hat { j } +N\hat { j } \) =0 (No acceleration)
By comparing the components on both sides, N-mg cos ፀ=0
N=mg cosፀ
The magnitude ~fnormal force (N) exerted by the surface is equivalent to mg cosፀ, Tlie object slides (with an acceleration) along the x direction. Applying Newton's second law in the x direction.
\(mg\ sin\theta\ \hat { i } =ma\hat { i } \)
By comparing the components on both sides, we can equate
mg sinፀ = ma
The acceleration of the sliding object is
a = g sinፀ

Note that the acceleration depends on the angle of inclination ፀ.

The gravitational potential energy (U) at some height his, equal to the amount of work required to take the object from the ground to that height h.
\(U=\int{\bar{F}_a.d\bar{r}}=\int_{0}^{h}|\bar{F}_a||d\bar{r}|\cos\ \theta\)
Since the displacement and the applied force are in the same upward direction, the angle between them, \(\theta = 0°.\)Hence, cos°=1 and \(|\bar{F}_a|=mg\) and \(|d\bar{r}|=dr.\)
\(U=mg\int_{0}^{4}dr\Rightarrow mg {[r]}_{0}^{h} =mgh\)

Consider the horse as the 'system', then there are three forces acting on the horse
(i) Downward gravitational force (m_hg)
(ii) Force exerted by the road (F_r)
(iii) Backward force exerted by the cart (F_c)
It is shown in the following figure.
F_r - Force exerted by the road on the horse
F_c - force exerted by the cart on the horse
F_r^丄 - Perpendicular component of F_r = N
F_r^||-Parallel component of F, which is reason for forward movement.

The force exerted by .the road can be resolved into parallel and perpendicular components, The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force (F_c). So there is net force along the forward direction which causes the forward movement of the horse.
If we take the cart as the system, then there are three forces acting on the cart.
(i) Downward gravitational force (m_cg)
(ii) Force exerted by the road (F_r')
(iii) Force exerted by the horse (F_h)

It is shown in the figure
The force exerted by the road (\(\vec { { F }_{ r } } \)) can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity (m_cg)
Parallel component acts backwards and the force exerted by the horse (\(\vec { { F }_{ h } } \)) acts forward. Force (\(\vec { { F }_{ h } } \)) is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward.
If we take the cart + horse as a system, then there are two forces acting on the system.
(i) Downward gravitational force (m_h + m_c)g
(ii) The force exerted by the road (F_r) on the system.
It is shown in the following figure.

(iii) In this case the force exerted by the road (F_r) on the system (cart + horse) is resolved in to parallel and perpendicular components. The perpendicular component is the normal force which cancels the downward gravitational force (m_h +m_c)g. The parallel component of the force is not balanced, hence the system (cart + horse) accelerates and moves forward due to this force.

(i) Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
(ii) Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
(iii) Draw the free body diagram.
(iv) Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the .system in vector notation and equate it to the right hand side of equation which is the product of mass .and acceleration. Here, acceleration should also be in vector notation.
(v) If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.
By following the above steps: We fix the inertial coordinate system on the. ground as shown in the figure.

The forces acting on the mango are
(i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis.
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis.
The free body diagram for the mango is shown in the figure.
\(\vec { { F }_{ g } } =mg(-\hat { j } )=-mg\hat { j } \)
Here, mg is the magnitude of the gravitational force and \((-\hat { j } )\) represents the unit vector in negative y direction.
\(\vec { T } =T\hat { j } \)



Here T is the magnitude of the tension force and \((-\hat { j } )\) represents the unit vector in positive y direction.
\({ \vec { F } }_{ net }={ \vec { F } }_{ s }+{ \vec { T } }_{ g }=-mg\hat { j } +T\hat { j } =(T-mg)\hat { j } \)
From Newton's second law \({ \vec { F } }_{ net }=m\vec { a } \)
Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero (\(\vec { a } =0\))
So \({ \vec { F } }_{ net }=m\vec { a } =0\)
\((T-mg)\hat { j } =0\)
By comparing the components on both sides of the above equation, we get T - mg = 0
So the tension force acting on the mango is given by T - mg
Mass of the mango m = 400g and g = 9.8 ms^-2 Tension acting on the mango is T = 0.4\(\times\)9.8 = 3.92 N.

Suppose, they are stretched by same distance x₁ and x₂ by the same force F.
Then, F = k, x₁ = k₂ x₂
\({W_1 \over W_2}={{{1\over 2}k_1{x}_{1}^{2}}\over{{1 \over 2}k_2{x}_{2}^{2}}}={{k_1.x_1.x_1}\over{k_2.x_2.x_2}}={F.x_1\over F.x_2}\)
\({W_1\over W_2}={x_1 \over x_2}\)  ....(1)
k₁x₁ = k₂x₂
\({{x_1}\over{x_2}}={k_2\over k_1}\)
Equation (1) becomes,   \({W_1 \over W_2}={k_2 \over k_1}\)
since, k₁ > k₂  W₁ > W₂
Suppose strings are stretched by same distance.
\({W_1 \over W_2}={{{1\over 2}k_1x^2}\over{{1\over 2}k_2}x^2}={k_1 \over k_2}\)
since k₁ > k₂  W₁ > W₂

(i) If the object is at rest and no external force is applied on the object, the static friction acting on the object is zero (f_s = 0).
(ii) If the object is at rest, and there is an external force applied parallel to the surface, then the force of static friction acting on the object is exactly equal to the external force applied on the object (f_s = F_ext). But still the static friction Is is less than μ_sN.
(iii) When object begins to slide, the static friction (f_s) acting on the object attains maximum.

If a force (F) acts on the object in a very short interval of time (M), from Newton's second law in magnitude form
Fdt = dp
Integrating over time from an initial time t_i to a final time t_f, we get
\(\int _{ i }^{ f }{ dp } =\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pf-pi = \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pi = initial momentum of the object at time t_i
P_t = final momentum of the object at time t_f.
pf - pi = Δp change in momentum of the object during the time interval
tf - ti = Δt
The integral \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)=J is called the impulse and it is equal to change in momentum of the object.
If the force is constant over the time interval, then
\(\int _{ { t }_{ i } }^{ { t }_{ f } }{ F } dt=\int _{ i }^{ f }{ dp } \) = F(t_f - t_i) = FΔt
FΔt = Δp

Galileo's experiment with. the second plane (a) at same inclination angle Cisthe first (b) with increased smoothness (c) with reduced angle of inclination (d) with zero angle of inclination
When a ball rolls from the top of an inclined plane to its bottom, after reaching the ground it moves some distance and continues to move on to another inclined plane of same angle of inclination as shown in the Figure (a). By increasing the smoothness of both the inclined planes, the ball reach almost the same height (h) from where it was released (L1) in the second plane (L2) [figure (b)]. The motion of the ball is then observed by varying the angle of inclination of the second plane keeping the same smoothness. If the angle of inclination is reduced, the ball travels longer distance in the second plane to reach the same height [figure (c)). When the angle of inclination is made zero, the ball moves forever in the horizontal direction [figure (d)]. If the Aristotelian idea were true, the ball would not have moved in the second plane even if its smoothness is made maximum since no force acted on it in the horizontal direction.

Force, F = 1000 N
Displacement, s = 200 m
Angle e = 60°
Workdone =?
Workdone W = Fs cos ,
= 1000\(\times\)200\(\times\)cos 60°
= 1000\(\times\)200\(\times\)\(\frac{1}{2}\)
=1\(\times\)10⁵ J = 1\(\times\)10² kJ

Mass of a body(m) =\(\frac{500}{1000}\)kg = 0.5 kg
Horizontal Force (F) = 1N
Time(t) = 20s.
Therefore, acceleration of a body.
a = F/m =\(\frac{1}{0.5}=2ms^{-2}\)
Distance travelled s = ut+\(\frac{1}{2}+at^{2}\)
S =0 x 20 +\(\frac{1}{2}\times2\times(20)^{2}\)
= 0+\(\frac{1}{2}\times400\times2\)
Displacement s¹= 400 m
Workdone = F\(\times\)s = 1\(\times\)400
\(\boxed{=400}\) J; v = u+at = 0+2\(\times\)20
v = 40
Change in k.E =\(\frac{1}{2}m(v^{2}-u^{2})\)
=\(\frac{1}{2}\times0.5\times(40^{2}-0)\)
=\(=\frac{1}{2}\times 0.5\times 1600\)
\(\boxed{=400\ J}\)

Mass of a body(m) =\(\frac{600}{1000}\)kg=0.6 kg
Velocity (v) = ax \(\frac{3}{2}\) where a = 3m^-1/4s^-1
We know that work done by net force acting on a body is equal to its change in k.E. If velocity of body corresponding to x = D and x = 3m be u and v respectively than 
Workdone W= \(\frac{1}{2}mv^{2}-\frac{1}{2}mv^{2}=\frac{1}{2}m(v^{2}-u^{2})\)
 Initial velocityu = a.(0)3/2= 0 and v = a.(3)^3/4
\(W=\frac { 1 }{ 2 } \times 0.6\left[ \left\{ a{ (3) }^{ \frac { 3 }{ 2 } } \right\} ^{ 2 }-0 \right] \)
=\(\frac{1}{2}\times0.6\times a^{2}\times 3^{\frac{3}{2}}\)
=\(\frac{1}{2}\times 0.6\times 3^{2}\times (3)^{3}\)
Workdone W = \(\frac{1}{2}\times 0.6 \times 9 \times 27 = 73J\)

displacement \(\vec { S } =(\overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } )\)m
Applied force \(\vec { F } =(2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +\overset { \wedge }{ 5k } )\)N
Angle (9) = 60°
Work done in moving a particle,
\(W=\vec(F).\vec{S}cos 60^{o}\)
=\((2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +\overset { \wedge }{ 5k } )\).\((\overset { \wedge }{ i } +2\overset { \wedge }{ j } +6\overset { \wedge }{ k } )\) cos 60^o
=\((\overset { \wedge }{ i } .\overset { \wedge }{ j }. \overset { \wedge }{ k } )(2+6+30)cos 60^{o}\)
W=38\(\times cos 60^{o}\)
=\(38\times\frac{1}{2}=19J \)  \([\because cos60^{o}=\frac{1}{2}]\)
Workdone W = 191.

The work done in moving particle.
\(W=\vec{F}.\vec{s}\)
\(\vec { F } =(2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )\)
\(\vec { s } =(\overset { \wedge }{ i } -2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )\)
Work done \(W=\vec{F}.\vec{s}\)
=\((2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } ).(\overset { \wedge }{ i } -2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )\)
=\((2\overset { \wedge }{ i } +6\overset { \wedge }{ j } +12\overset { \wedge }{ k } )\)
=\((\overset { \wedge }{ i } .\overset { \wedge }{ j } .\overset { \wedge }{ k } ).(2+6+12)\)
=1.(2+6+12)
Work done(W) = (20)J.

Mass of the bullet (m) = 25 g
Initial velocity (m) = 400 ms^-1
Final velocity (v) = 300 ms^-1
Work done (w) =?
From work-energy theorem,
Work done = loss in kinetic energy,
\(W=\frac{1}{2}m(u^{2}-v^{2})\)
=\(\frac{1}{2}\times\frac{25}{100}kg\times(400^{2}-300^{2})\)
=\(\frac{1}{2}\times0.025\times(16\times10^{4}-9\times10^{4})\)
=\(\frac{1}{2}\times0.025\times(7\times10^{4})\)
= 0.0125\(\times(7\times10^{4})\)
Work done (W) = 875J

Here the motion is not accelerated the resultant force parallel to the plane must be zero.

So,
F = Mg sin 45° = O.
Force, F = Mg sin 45°;
distance d = 10m
the workdone
W = Fd cos,
= (Mg sin 45°)d cos0° [ ∵ θ = 0° so, cos 0° = 1]
= 20\(\times\)10\(\times\)sin 45°\(\times\)10\(\times\)cos 0°
= 20\(\times\)10\(\times\)\(\frac{1}{\sqrt{2}}\)x 10\(\times\)1
= 1414J
Workdone W = 1414J

\(\vec { { r }_{ 1 } } =(2\overset { \wedge }{ i } +\overset { \wedge }{ j } -\overset { \wedge }{ 3k } )\)
\(\vec { { r }_{ 2 } } =(4\overset { \wedge }{ i } +6\overset { \wedge }{ j } -7\overset { \wedge }{ k } )\)
The position vectors \(\vec{r}=\vec{r_2}-\vec{r_1}\)
=\(\left( 4\overset { \wedge }{ i } +\overset { \wedge }{ 6 } j-\overset { \wedge }{ 7k } \right) -\left( \overset { \wedge }{ 2i } +\overset { \wedge }{ j } -\overset { \wedge }{ 3k } \right) \)
\(=\overset { \wedge }{ 4i } +\overset { \wedge }{ 6j } -\overset { \wedge }{ 7k } -\overset { \wedge }{ 2i } -\overset { \wedge }{ j } +\overset { \wedge }{ 3k } \)
\(=\overset { \wedge }{ i } (4-2)+\overset { \wedge }{ j } (6-1)+\overset { \wedge }{ k } (-7+3)\)
\(=\overset { \wedge }{ i } (2)+\overset { \wedge }{ j } (5)+\overset { \wedge }{ k } (-4)\)
\(\vec { r } =\vec { { r }_{ 2 } } -\vec { { r }_{ 1 } } =2\overset { \wedge }{ i } +5\overset { \wedge }{ j } -\overset { \wedge }{ 4k } \)
The effect of force  = \(\vec { F } =(3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ 4k } )N\)
Workdone W = \(\vec{F}.\vec{x}\)
\(=(3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ 4k } ).(2\overset { \wedge }{ i } +5\overset { \wedge }{ j } -\overset { \wedge }{ 4k } )\)
\(=\overset { \wedge }{ i } .\overset { \wedge }{ j } .\overset { \wedge }{ k } (6+10-16)\)
= 1(0)
Work done = 0.

Applied force = 10 N
Opposing friction forcej= M_k. N = M_k·mg.
= 0.3\(\times\)3\(\times\)9.8 = 8.82 N.
Net accelerating forceF - f = 10N - 8.82N
=1.18N
Acceleration a =\(\frac{force}{mass}\)=\(\frac{8.82N}{3Kg}=2.94 ms^{-2}\)
Distance covered in 10s (assuming w = 0)
\(s=0+\frac{1}{2}at^{2}=\frac{1}{2}\times2.94\times(10^{2}) = 147 m\)
there force workdone by a applied force,
W = Fs = 10\(\times\)147
W = 1470 J

(i) For a constant force, the impulse is denoted as \(j=F\triangle t\) and it is also equal to change in momentum \((\triangle p)\)  of the object over the time interval \(\triangle t\)
Impulse is a vector quantity and its unit is Ns.
(ii) The average force acted on the object over the short interval of time is defined by
\({ F }_{ avg }=\frac { \triangle p }{ \triangle t } \)           ..........(1)       
(iii) From equation (1), the average force that act on the object is greater if t is smaller. Whenever the momentum of the body changes very quickly, the average force becomes larger.
(iv) The impulse can also be written in terms of the average force. Since \(\triangle\) p is change in momentum of the object and is equal to impulse (J), we have
\(j={ F }_{ avg }\triangle t\)       .........(2)
(v) The graphical representation of constant force impulse and variable force impulse is given in Figure.

Work-Kinetic Energy Theorem
Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.
The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs
The constant force is given by the equation,
F = ma
The third equation of motion can be written as,
\(v^{2} =u^{2}+2 a s  \)
\(a =\frac{v^{2}-u^{2}}{2 s}\)
Substituting for a in equation (2),
\(F=m\left(\frac{v^{2}-u^{2}}{2 s}\right)\)
Substituting equation (2), (1)
\(w=m\left(\frac{v^{2}}{2 s} s\right)-m\left(\frac{u^{2}}{2 s} s\right)  \)
\(w=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
The expression for kinetic energy:
The term \(\left(\frac{1}{2} m v^{2}\right)\) in the above equation is the kinetic energy of the body of mass (m) moving with velocity(v).
\(K E=\frac{1}{2} m v^{2}\)
Kinetic energy of the body is always positive. From equations (4) and (5)
\(\Delta K E =\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}  \)
\(\text {Thus, } W =\Delta K E\)
The expression on the right hand side (RHS) of equation (6) is the change in kinetic energy (\(\Delta\)KE) of the body.
This implies that the work done by the force on the body changes the kinetic energy of the body, This is called work-kinetic energy theorem.
The work-kinetic energy theorem implies the following.
1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

(i) Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest).
(ii) If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.
(iii) This implies that in addition to the inward centripetal force - mω²r there must be an equal and opposite force that acts on the stone outward with value + mω²r.
(iv) So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω²r=0).
(v) This outward force + mω²r is called the centrifugal force.

The force which always opposes the relative motion between an object and the surface where it is placed is called frictional force. Frictional force always acts on the object parallel to the surface on which the object is placed. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But now it is understood that frictional force is the electromagnetic force between the atoms on the two surfaces. Because of this even a well polished surfaces have frictional force. Frictional force is independent of surface area but depends on the normal force acting on it.
Angle of friction:
The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (f_s^max). from the figure, tan ፀ \(=\frac{f_s^{max}}{N}\)

But f_s^max = μs N where us is the coefficient of static friction.
∴ μs = tan ፀ
To show that angle of friction is equal to angle of repose

After collision, along x-axis
\(m_{1} u_{1}=m_{1} v_{1} \cos \theta_{1}+m_{2} v_{2} \cos \theta_{2} \)
Along Y-axis
\(0 =m_{1} v_{1} \sin \theta_{1}-m_{2} v_{2} \sin \theta_{2}  \)
\(m_{1} v_{1} \sin \theta_{1} =m_{2} v_{2} \sin \theta_{2}  \)
\(m_{1} =m_{2}  \)
\(\therefore v \sin \theta =\frac{v}{2} \sin 90^{\circ}  \)
\(\sin \theta =\frac{1}{2} \times 1=\frac{1}{2}  \)
\(\therefore \theta =\sin ^{-1}(0.5)=30^{\circ}\)

Radius (r) of moon orbit from the centre of earth
\(r =(384,000 \mathrm{~km}) \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \)
= 384,000,000 m
Time Period (T) = (27 days ) \(\times \frac{24 \text { hours }}{1 \text { day }} \times \frac{60 \mathrm{~min}}{1 \mathrm{hr}} \times \frac{60 \mathrm{sec}}{1 \mathrm{~min}}\)
= 2,332,800 sec
\(velocity (v)=\frac{\text { circumference }}{\text { Time period }}=\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)
\(=\frac{2 \pi \times 384,000,000 m}{2,332,800 \mathrm{~s}}=329 \pi \mathrm{m} / \mathrm{s}\)
Centripetal acceleration \(\left(a_{\perp}\right)=\frac{v^{2}}{r}=\frac{(329 \pi \mathrm{m} / \mathrm{s})^{2}}{384,000,000 \mathrm{~m}}\)
\(a_{\perp}=2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}\)
Tension due to Centripetal force
\(F_{c} =T=m a_{\perp}=7.34 \times 10^{22} \mathrm{~kg} \times 2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2} \)
\(\mathrm{~T} =2.04052 \times 10^{20} N\)

Force acting vertically to sling shot
Tension T in each string;
Now applying Lami's theorem, we get
\(\frac{2 T}{\sin \theta} =F  \)
\(\frac{2 T}{\sin (2 \times 30)} =50 N  \)
\(\frac{2 T}{\sin (60)} =50 N  \)
\(\frac{2 T}{\sqrt{3}} =50 N  \)
\(\frac{T}{2 T} =\frac{2}{\sqrt{3}} \times 50 N  \)
\(T=\frac{50 N}{\sqrt{3}}=28.268 N\)

For the system to be in equilibrium for block of mass M1 frictionless Tension in the string T
T = μ_s (M₃ + M₁)g
For the equilibrium of block of mass M₂
T = M₂g .....(2)
(1) = (2)
μ_s (M₃ + M₁)g = M₂g
M₃ =\(\frac { { m }_{ 2 } }{ { \mu }_{ 2 } } -{ m }_{ 1 }\) 

\(F_{1 x} =F_{1} \cos (30)  \)
\(=50 \times \frac{\sqrt{3}}{2}  \)
\(m =20 \mathrm{~kg}  \)
\(\frac{F_{1 x}}{m} =a_{x} \Rightarrow a_{x}=\frac{25 \times \sqrt{3}}{20}=2.165 \mathrm{~ms}^{-1}  \)
\(F_{1 y} =F_{1} \sin (30)=50 \times \frac{1}{2}=25  \)
\(\frac{F_{1 y}}{m} =a_{y} \Rightarrow a_{y}=\frac{25}{20}=1.25 \mathrm{~ms}^{-1}\)
 

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

Final values of potential energy, kinetic energy and total energy are measured at the height h.
By law of conservation of energy, the initial and final total energies are the same.
\(\frac { 1 }{ 2 } { mv }_{ 0 }^{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }+mgh\)
\({ v }_{ 0 }^{ 2 }={ v }^{ 2 }+2gh\)
\(v=\sqrt { { v }_{ 0 }^{ 2 }-2gh } \)

(a) The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At h = 10 m, the total energy E is entirely potential energy.
E = U = mgh = 1 × 10  × 10 = 100 J
(b) The potential energy of the object at h = 4 m is
U = -mgh = 1 \(\times\) 10 \(\times\) 4 = 40 J
(c) Since the total energy is constant throughout the motion, the kinetic energy at h = 4m must be KE = E - U = 100 - 40 = 60 J.
Alternatively, the kinetic energy could also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m.
The velocity after falling 6 m is calculated from the equation of motion,
\(v=\sqrt { 2gh } =\sqrt { 2\times 10\times 6 } =\sqrt { 120 } \) ms^-1
v² = 120
The kinetic energy is KE = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \times 1\times 120=60J\)
(d) When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy, U = 0.
\(E=KE=\frac { 1 }{ 2 } { mv }^{ 2 }=100\quad J\)
\(v=\sqrt { \frac { 2 }{ m } KE } =\sqrt { \frac { 2 }{ 1 } \times 100 } =\sqrt { 200 } =14.12\) ms^-1

The spring constant, k = 0.1 N m^-1
The displacement, x = 25 cm = 0.25 m
(a) The potential energy stored in the spring is given by
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=0.0031J\)
(b) The work done W_s by the spring force _{\(\bar { F } \)} is given by,
\({ W }_{ s }=\int _{ 0 }^{ x }{ \overrightarrow { { F }_{ s } } .\overrightarrow { dr } } =\int _{ 0 }^{ x }{ (-k\ x\hat { i } ).(dx\hat { i } ) } \)
The spring force \(\overrightarrow { { F }_{ s } } \)  acts in the negative x direction while elongation acts in the positive x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } \)
\({ W }_{ s }=-\frac { 1 }{ 2 } \times 0.1\times { (0.25) }^{ 2 }=-0.0031\ J\)
Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.
(c) During compression also the potential energy stored in the object is the same.
\(U=\frac { 1 }{ 2 } { kx }^{ 2 }=0.0031\ J\)
Work done by the restoring spring force during compression is given by
\({ W }_{ s }=\int _{ 0 }^{ x }{ { \overrightarrow { F } }_{ s } } \overrightarrow { dr } =\int _{ 0 }^{ x }{ (kx\hat { i } ).(-dx\hat { i } ) } \)
In the case of compression, the restoring spring force acts towards positive x-axis and displacement is along negative x direction.
\({ W }_{ s }=\int _{ 0 }^{ x }{ (-kx)dx=-\frac { 1 }{ 2 } { kx }^{ 2 } } =-0.0031J\)

When the object goes up, the displacement points in the upward direction whereas the gravitational force acting on the object points in downward direction. Therefore, the angle between gravitational force and displacement of the object is 180°.
(a) The work done by gravitational force in the upward motion.
Given that dr = 5m and F = mg
W_up = Fdr cosθ = mgdr cos 180°
W_up = 2\(\times\)10\(\times\)5\(\times\) (-1) = -100 joule
[cos 180° = -1]
(b) When the object falls back, both the gravitational force and displacement of the object are in the same direction. This implies that the angle between gravitational force an displacement of the object is 0°.
W_down = Fdr cos 0°
W_down = 2\(\times\)10\(\times\)5\(\times\)(1) = 100 joule
[cos 0° = -1]
(c) The total work done by gravity in the entire trip (upward and downward motion)
W_total,= W_up +W_down
= -100 joule + 100 joule = 0

Mass of a bob = m
Length of the rod = r
From the law of conservation of energy,
\(\frac{1}{2} m v_{1}^{2} =2 m g r+\frac{1}{2} m v_{2}^{2}  \)
\(\therefore \frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2}\right) =2 m g r  \)
\(\therefore v_{1}^{2}-v_{2}^{2} =4 g r  \)
\(\text { If } v_{2} =0 \text { then }  \)
\(v_{1}^{2} =4 g r  \)
\(\therefore v_{1} =\sqrt{4 g r} m s^{-1}\)

\(\text {Centrifugal force experience by person } =\frac{m v^{2}}{r}=\frac{60 \mathrm{~kg} \times(50 \mathrm{~m} / \mathrm{s})^{2}}{10 \mathrm{~m}}  \)
\(=\frac{60 \times 2500}{10} \mathrm{~N}  \)
\(=15000 \mathrm{~N}\)

\(F-T =m_{1} a \ \& \ T=m_{2} a  \)
\(a =\frac{F}{m_{1}+m_{2}} \ \& \ T=\frac{m_{2} F}{m_{1}+m_{2}}  \)
\(T =\frac{10 k g \times 500 N}{(10 k g+15 k g)}  \)
\(T =\frac{5000}{25} N  \)
\(T =200 \mathrm{~N}\)

Relation between power and velocity
The work done by a force \(\overrightarrow{\mathbf{F}}\) for a displacement \(d \vec{r}\) is
\(W=\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}\)  ......(1)
Left hand side of the equation (1) can be written as
\(W=\int d W=\int \frac{d W}{d t} d t\)
(multiplied and divided by dt) (2)
Since, velocity is \(\vec{v}=\frac{d \vec{r}}{d t} ; \overrightarrow{d r}=\vec{v} d t.\) Right hand side of the equation (1) can be written as
\(\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}=\int\left(\overrightarrow{\mathrm{F}}, \frac{d \vec{r}}{d t}\right) d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\left[v=\frac{d \vec{r}}{d t}\right] \ldots \ldots\) (3)
Substituting equation (2) and equation (3) in equation (1), we get
\(\int \frac{d W}{d t} d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\)
Or
\(\int\left(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v}\right) d t=0\)
This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v} =0  \)
\(\frac{d W}{d t} =\overrightarrow{\mathbf{F}} \vec{v}\)
Examples: Motors, Engines and Automobiles
A vehicle of mass 1250 kg is driven with an acceleration 0.2 ms^-2 along a straight level road against an external resistive force 500 N. Calculate the power delivered by the vehicle's engine if the velocity of the vehicle is 30 ms^-1.
Solution
The vehicle's engine has to do work against resistive force and make vehicle to move with an acceleration. Therefore, power delivered by the vehicle engine is
\(P =(\text { resistive force }+\text { mass } \times \text { acceleration }) \text { (velocity) }  \)
\(P =\overrightarrow{\mathbf{F}}_{-\mathrm{ma}} \vec{v}=\left(F_{\text {reistie }}+F\right) \vec{v}  \)
\(P =\overrightarrow{\mathbf{F}}_{\text {tot }} \vec{v}=\left(F_{\text {reithie }}+m a\right) \vec{v}  \)
\(=500 \mathrm{~N}+\left((1250 \mathrm{~kg}) \times\left(0.2 \mathrm{~ms}^{-2}\right)\right)\left(30 \mathrm{~ms}^{-1}\right)=22.5 \mathrm{kw}\)

(i) Tangential acceleration = g sin ፀ
(ii) mg cos ፀ acts along OP (outwards)
(iii) tension (T) acts along PO (inwards)
Net force on the body at P acting along
PO = T- mg cos ፀ

This must provide the necessary centripetal force. \(\frac{mv^2}{r}\)
\(\therefore T -mg \ cos \ \theta=\frac{mv^2}{r} \)  
Centripetal acceleration (a_⊥) =  \(\frac { T-mg \ cos\theta }{ m } \)