For a conservative force
\(F=-\frac{d v}{d t} \)
\(\int_{0}^{u(x)} d v=-\int_{0}^{x} F d x=-\int_{0}^{x} k x d x \)
\(\text { As } v(0)=0\)
\(U(x)=-\frac{k x^{2}}{2}\)
Thus, the graph of U(x) versus (x) will be a parabola, symmetric about U - ax is bying below x - ax is with its vertex at the origin. Hence the correct answer is C

\(\text {Potential energy } P . E=\alpha-\frac{\beta}{2} x^{2}\)
P.E = Work = Fx
\(P=\alpha-\frac{\beta}{2} x^{2}\)
\(\text {Force }=\frac{d p}{d x}=\frac{d}{d x}\left(\alpha-\frac{\beta}{2} x^{2}\right)\)
\(=0-\frac{\beta}{2} \times 2 x =-\beta x \)

Radius =R
\(v_{1}^{2}-v_{2}^{2}=4 g R\)
\(\text { Tension } T_{2}=\frac{m v_{2}^{2}}{R_{2}}-m g\)
\(\text {To find minimum speed, let } T_{2}=0\)
\(0 =\frac{m v_{2}^{2}}{R}-m g \)
\(\frac{m v^{2}}{R} =m g \)
\(v_{2}^{2}=R g \ v_{2} =\sqrt{g R} \)
\(\text { sub (2) in the eqn (1) we get }\)
\(v_{1}^{2}-(\sqrt{g R})^{2} =4 g R \)
\(v_{1}^{2}-g R =4 g R \)
\(v_{1}^{2} =4 g R+g R \)
\(=5 g R \)
\(v_{1} =\sqrt{5 g R} \)

Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

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

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

Consider a spring-mass system. Let us assume a mass, m lying on a smooth horizontal table as shown in the figure. Here, x = 0 is the equilibrium position. One end of the spring is attached to a rigid wall and the other end to the mass.
As long as the spring remains in equilibrium position, its potential energy is zero. Now an, external force \(\bar{F},\) is applied 'so that it is stretched by a distance (x) in the direction of the force.
There is a restoring force called spring force Fs developed in the spring which tries to bring the mass back to its original position. This applied force and the spring force are equal in magnitude but opposite in direction i.e., \(\bar{F}_a=-\bar{F}_s.\) 

According to Hooke's law, the, restoring force developed in the spring is
\(\bar{F}_s=-k.\bar{x}\)    ...(1)
The negative sign 'in the above expression implies that the spring force is always opposite to that of displacement i and k is the force constant. Therefore applied force is \(\bar{F}_a=+k.\bar{x}\). The positive sign implies that the applied force is in the direction of displacement \(\bar{x}\). The spring force is an example of variable force as it depends on the displacement \(\bar{x}\). Let the spring be stretched to a small distance d \(\bar{x}\). The work done by the applied force on the spring to stretch it by a displacement x is stored as elastic potential energy.
\(U=\int{\bar{F}_ad\bar{r}}=\int_{0}^{\pi}|\bar{F a}||d\bar{r}|\cos\theta\)
\(U=\int_{0}^{x}F_adx \cos \theta\)   ...(2)
The applied force \(\bar{F}_a\) and the displacement \(d\bar{r}\) (i.e., here dx) are in the same direction. As, the initial position is taken as the equilibrium position or mean position, x = 0 is the lower limit of integration.
\(U=\int_{0}^{\pi}kxdx\)    ........(3)
\(U=k{\left[ {x^2 \over 2} \right]}_{x}^{0}\) ............(4)
\(U={1\over2}kx^2\)    ........(5)
If the initial position is not zero, and if the mass is changed from position x_i to x_f, then the elastic potential energy is
\(U={1\over2}k({x}_{f}^{2}-{x}_{i}^{2})\)  .......(6)
From equations (5) and (6), we observe that the potential energy of the stretched spring depends on the force constant k and elongation or compression x.

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.

v₁ = 100 m/s
s₁ =4 cm
s₂ =9 cm
v₂ =??
K.E lost = Word done against plank's resistance.
\({1\over 2}{mv}_{1}^{2}=F\times s_1\)
\({1\over 2}{mv}_{1}^{2}=F\times s_2 \)
\({(2)\over(2)}\Rightarrow{{v}_{2}^{2} \over {v}_{1}^{2}}={s_2\over s_1}\Rightarrow{v_2 \over v_2}=\sqrt{{s_2 \over s_1}}\)
\({v_2 \over v_1}=\sqrt{{9 \over 4}}={3 \over 2}\Rightarrow v_2={3 \over 2}\times v_1\)
\(v_2={3 \over 2}\times100=150m/s\)

(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.

m₁ = 10kg, m₂ = 20kg
u₁=20ms^-1, u₂ = -10 ms^-1
\(v_1={{m_1-m_2}\over{m_1+m_2}}.u_1+{{2m_2}\over{m_1+m_2}}.{u}_{2}\)
\(={2\times 10\over10+20}\times20+{2\times 20\over 10+20}\times(-10)\)
\(={-20\over3}-{40\over3}={-60\over3}=-20{ms}^{-1}\)
\(v_2={2m_1\over m_1+m_2}u_1+{m_2-m_1\over m_2-m_2}.u_2\)
\(={20\times10\over10+20}\times20+{20-10\over10+20}-(-10)\)
\(={40\over3}-{10\over3}={30 \over3}=10{ms}^{-1}\)

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.

(i) A body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius \(\vec{r}\) of the circular path.
(ii) The motion of the body by taking the free body diagram (FBD) at a position where the position vector \(\vec{r}\) makes an angle e with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, In the tangential direction,

mg sinθ = ma_t
mg sinθ = -m \((\frac{dv}{dt})\)
where, a_t = -\((\frac{dv}{dt})\) is tangential retardation
In the radial direction,
T - mg cosθ = m a_r
T - mg cosθ = \(\frac{mv^{2}}{r}\)
where, a_r = \(\frac{v^{2}}{r}\) is the centripetal acceleration.

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.

i. Every object continues to be in the state of rest or of uniform motion unless there is external force acting on it.
ii. The force acting on an object is equal to the rate of change of its momentum.
iii. For every action there is an equal and opposite reaction.
Discussion:
i) Newton's laws are vector laws. The equation \(\vec{F}=\mathrm{m} \vec{a}\) can be written in cartesian coordinates as
\(\mathrm{F}_{x} \hat{\mathrm{i}}+\mathrm{f}_{\mathrm{y}} \hat{\mathrm{j}}+\mathrm{F}_{z} \hat{\mathrm{k}}=\operatorname{ma}_{x} \hat{\mathrm{i}}+\operatorname{ma}_{\mathrm{y}} \hat{\mathrm{j}}+\operatorname{ma}_{\mathrm{z}} \hat{\mathrm{k}}\)
On comparing both sides,
\(\mathrm{F}_{x}=m \mathrm{~m}_{x}  \)
\(\mathrm{F}_{\mathrm{y}}=m \mathrm{ma}_{\mathrm{y}}  \)
\(\mathrm{F}_{\mathrm{z}}=m \mathrm{ma}_{\mathrm{z}}\)
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along n direction. In the same way, F₂ can not affect a_y and a_n
ii) The acceleration experienced by the body at a time t depends on the force which acts on the body at that instant of time. Thus, \(\vec{F}(\mathrm{t})=\mathrm{m} \vec{a}(\mathrm{t})\)
when a bowler throws the ball to a batsman the acceleration of the ball is determined by the gravitational and air frictional forces and not by the speed which it is thrown.
iii) The direction of force may be different from the direction of force. The following motions are possible.
a) Force and motion in the same direction.
Example: Falling of an apple from the tree.
b) Force and motion are not in the same direction.
Example: The Moon experiences a force towards the Earth but Moon moves in the elliptical path.
c) Force and motion are in the opposite direction.
Example: The motion of a body thrown vertically upward.
d) Zero net force, but there is motion.
Example: The falling of rain drops.
4) If multiple forces \(\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3} \ldots\) act on the same body then total force is equal to the vectorial sum of the individual forces \(\vec{F}_{\text {net }}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\ldots\)
5) Newton's second law can be written in the second derivative of position vector as  \(\vec{F}=\mathrm{m} \frac{\mathrm{d}^{2}\overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}\). It means that whenever the second derivative of position vector is not zero, there must be a force acting on the body.
6) If no force acts on the body then \(m \frac{d \vec{v}}{\mathrm{dt}}=0\). It implies that \(\vec{v}\) is constant. Thus second law is consistent with the first law even though they are independent to each other.
7) Newton's second law is cause and effect relation force is the cause and acceleration is effect.

30% of Power=\(\frac{W}{t}=\frac{mgh}{t}=\frac{V\rho gh}{t}\)
\(\frac{30}{100}\times P=\frac{V\rho gh}{t}\)
P=43.6 KW

P = 8\(\times\)(kinetic energy of each bullet per second)²
= 8\(\times\)\(\frac{1}{2}\) \(\times\)(3\(\times\)10^-3)\(\times\)(600)²
P = 4320W
P = 4.320 kW

At maximum compression x_m the K.E. of the car is converted entirely into the P.E. of the spring.
\(\therefore\) \(\frac { 1 }{ 2 } k{ x }_{ m }^{ 2 }=\frac { 1 }{ 2 } m{ v }^{ 2 }or{ \ x }_{ m }=2m\)

The downward force on the elevator is :
F = mg + f = 22000 N
\(\therefore\) Power supplied by motor to balance this force is:
P = Fv = 44000 W =\(\frac { 44000 }{ 746 } \)= 59 hp.

For x = 10 cm = 0·1 m, Fx =\(\frac { 1 }{ 2 } m{ v }_{ 1 }^{ 2 }=1J\)
\(\therefore\)  F = 10N
For x=6cm =0.06m, Fx = \(\frac { 1 }{ 2 } m{ v }_{ 1 }^{ 2 }-\frac { 1 }{ 2 } m{ v }_{ 2 }^{ 2 }\)
or Fx = \(\frac { 1 }{ 2 } m{ v }_{ 1 }^{ 2 }-\)Final K.E.
or Final K.E.= \(\frac { 1 }{ 2 } m{ v }_{ 1 }^{ 2 }-\) Fx = 1-10\(\times\)0.06 = 1 - 0.6 = 0.4 J

Use F = ma
\(a=\frac { v-u }{ t } =\frac { 200-0 }{ 4 } =50\ cm/s^{ 2 }\)
F = 500 \(\times\) 50 = 25,000 dyne.

\(h=ut+\frac { 1 }{ 2 } { gt }^{ 2 }\)
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg

As \(F\infty \frac { dP }{ dt } \)
when \(F=0,\frac { dP }{ dt } =0\)  so P = constant

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

We calculated in the previous chapter that the acceleration experienced by the object in the inclined plane as g sinθ.
According to Newton's second law, the force acting on the mass along the inclined plane F = mg sinθ. Note that this force is constant throughout the motion of the mass. 
The work done by the parallel component of gravitational force (mg sin θ) is given by
W = \(\vec{F}.d\vec{r}\) = Fdr cosФ
where Ф is the angle between the force (mg sin θ) and the direction of motion (dr). In this case, force (mg sin θ) and the displacement (d \(\vec r\)) are in the same direction. Hence Ф = 0 and cos Ф = 1.
W = F dr = (mg sin θ) (dr)
(dr = length of the inclined place)
W = 1\(\times\)10\(\times\)sin (30°)\(\times\)10 = 100 \(\times\)\({1\over 2}=50J\)

(i) Tension T₁ of 24N pulls the masses (3m + 5m) with acceleration a.
∴ 24 = (3m + 5m)a or a = \(\frac { 3 }{ m } \)
(ii) Tension T₂ pulls mass 5m with acceleration \(\frac { 3 }{ m } \)
∴ T₂ = 5m\(\times\) \(\frac { 3 }{ m } \) = 15N
(iii) F = (m + 3m + 5m)a = 9m\(\times\)\(\frac { 3 }{ m } \) = 27N

Given: Acceleration = 4ms^-2
Time = 10s
(i) Horizontal velocity of the bus or the stone at t = 10s is
vx = u + at = 0 + 4\(\times\)10 = 40ms^-1
For vertical motion of the stone,
u = 0, a = g =10 ms^-2, t = 10.2 - 10 = 0.2s
∴ vy = 0 + 10\(\times\)0.2 = 2ms^-1
Magnitude of the resultant velocity of the stone is
v = \(v=\sqrt { { v }_{ x }^{ 2 }+{ v }_{ y }^{ 2 } } =\sqrt { { 40 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 1604 } \)
(ii) After the stone is dropped, its acceleration along horizontal is zero. It has only a vertical acceleration of 10 ms^-2

Yes, there is ail internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum of the momenta of the moving particles may be zero.

Both the coolies do the same amount of work, so they spend the same amount of energy.
W =P₁t₁ = P₂t₂
\({p_2 \over p_1}={t_1 \over t_2}={60\ sec \over 30\ sec}=2\)
(or) P₂ = 2P₁ 2^nd coolie has greater power.

(i) The acceleration is inversely proportional to mass. For the same force, the heavier mass experiences lesser acceleration and the lighter mass experiences greater acceleration.
(ii) When an apple falls, it experiences Earth's gravitational force. According to Newton's third law, the apple exerts equal and opposite force on the Earth.
(iii) Even though both the apple and Earth experience the same force, their acceleration is different.
(iv) The mass of Earth is enormous compared to that of an apple. So an apple experiences larger acceleration and the Earth experiences almost negligible. acceleration.
(v) Due to the negligible acceleration, Earth appears to be stationary when an apple falls.

Mass of a rocket, m = 7000 kg
Acceleration of the rocket, a = 30 m/s²
Speed of a rocket, v = 700 m/s
Force on the rocket, F = m(g + a)
F = m(g+a)   [\(\because\) g\(\rightarrow\)acceleration due to gravity is 9.8 ms^-2]
= 7000 (9.8 + 30)
= 7000(39.8)
F = 2.786\(\times\)10⁵N
Force, F v\(\frac { dm }{ dt } \)
\(\frac { dm }{ dt } =\frac { F }{ v } \)
= \(=\frac { 2.789\times { 10 }^{ 5 } }{ 700 } \)
Amount of gases ejected per second = 398 kgs^-1

Mass of a body m = 2 kg
Magnitude of the two force, F₁= 3N and F₂ =5N,
Angle (\(\theta\) ) = 90°
Resultant force on the body = 
F = \(\sqrt { { F }_{ 1 }^{ 2 }={ F }_{ 2 }^{ 2 }+2{ F }_{ 1 }{ F }_{ 2 }cos\theta } \)
= \(\sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 }+2\times 3\times 5\times cos90° } \)
= \(\sqrt { 9+25+2\times 3\times 5\times 0 } \)   \(\left[ \because \cos90°=0 \right] \)
= \(\sqrt { 9+25 } \)
F = \(\sqrt { 34 } \)
Net acceleration experienced by the body
a =\(\frac { F }{ m } \)
= \(\frac { \sqrt { 34 } }{ 3.5 } \) = 1.66 ms^-2

A force is said to be non-conservative if the work done by or against the force in moving a body depends upon the path between the initial and final positions. This means that the value of work done is different in different paths.
(i) Frictional forces are non-conservative forces as the work done against friction depends on the length of the path moved by the body.
(ii) The force due to air resistance, viscous force are also non-conservative forces as the work done by or against these forces depends upon the velocity of motion.

(i) Consider a force \(\vec {(F)}\) acting on a body which moves displacement in some direction\(​​\vec{(dr)}\) as shown in figure.
work done = force\(\times\)displacement, W = Fs
(ii) When force and displacement are inclined in each other, the expression for work done (W) by the force on the body is mathematically written as, \(W=\vec{F}=\vec{dr}\)
Here, the product \(\vec{F}.\vec{dr}\) is a scalar product (or dot product). Thus, work done is a scalar quantity.
w=F dr cosθ
where, θ is the angle between applied force and the displacement of the body.
The work done by the force depends on the force (F), displacement (dr) and the angle (θ) between them.

Forces on Block A:
(i) Downward gravitational force exerted by the Earth (m_Ag)
(ii) Upward normal force (N_B) exerted by block B (N_B)
The free body diagram for block A is as shown in the following picture.

Forces on Block B:
(i) The downward gravitational force exerted by Earth (m_Bg)
(ii) The downward force exerted by block A (N_A)
(iii) An upward normal force exerted by block C(N_c)

Forces on Block C:
(i) Downward gravitational force exerted by Earth (m_Cg)
(ii) Downward force exerted by block B (N_B)
(iii) Upward force exerted by the table (N_table)

We know that a=\(\left[ \frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } \right] =\frac { 50N }{ 10kg+7kg+2kg } =\frac { 50 }{ 19 } \)= 2.63 ms^-2

It is the restoring force set up in a string per unit extension.

The lighter body has more K.E. as K.E. = \(\frac { { p }^{ 2 } }{ 2m } \) and for constant p, K.E.\(\propto \frac { 1 }{ m } \) .

For motion, there should be external force.

Since the object is at rest, the gravitational force experienced by an object is balanced by normal force exerted by floor.
N = mg
The maximum static frictional force f_s^max = μ_sN = μ_sN = μ_smg
f_s^max= 0.8\(\times\)2 9.8 =15.68 N.
Therefore to move the object the external force should be equal to maximum static friction.
F_ext = 15.68 N.

Work done, \(W-\int^{x_f}_{x_i}F(x)dx=k\int_0^4x^2 dx={64\over 3}Nm\)

Force, F = 25 N
Displacement, dr = 15 m
Angle between F and dr, θ = 30°
Work done, W = F dr cos θ
W = 25 x 15 x cos 30° = 25\(\times\)15\(\times\) \(\sqrt3\over 2\)
W = 324.76 J

When the tyre is properly inflated, the area of contact between the tyre and the ground is reduced. This reduces rolling friction. Consequently, the automobile covers greater distance for the same quantity of fuel consumed.

1 electrical unit = 1 kWh = 1\(\times\)(103 W)\(\times\)(3600 s)
1 electrical unit = 3600\(\times\)103 W s
1 electrical unit = 3.6\(\times\)10⁶ J
1 kWh = 3.6\(\times\)10⁶ J

(i) Magnitude of the force.
(ii) Magnitude of the displacement of the body.
(iii) Angle between the force and displacement.

The energy is dissipated in the form of heat. The heat energy so produced is not available for work.

As P.E. = mgh \(\Rightarrow\) P.E., \(\infty\) h. So the graph of P.E. verses height is a straight line as shown.

The centripetal acceleration of a particle in the circular motion is given by a= \(\frac { { v }^{ 2 } }{ r } \) and it acts towards center of the circle. According to Newton's second law, the centripetal force is given by \({ F }_{ cp }={ ma }_{ cp }=\frac { mv^{ 2 } }{ r } \).
Eg: In the case of whirling motion of a stone tied to a string, the centripetal force on the particle is provided by the tensional force on the string. 

Frictional force which always opposes the relative motion between an object and the surface where it is placed.

Aristotelian law of motion states that "Force causes motion". This is wrong. It is based on common sense. In practise, force is required to counter the opposing force of friction.

(i) The graph shows that static friction increases linearly with external applied force till it reaches the maximum.
(ii) If the object begins to move then the kinetic friction is slightly lesser than the maximum static friction.
(iii) Note that the kinetic friction is constant and it is independent of applied force.

(i) If multiple forces \(\overset { \rightarrow }{ { F } } _{ 1 },\overset { \rightarrow }{ { F } } _{ 2 },\overset { \rightarrow }{ { F } } _{ 3 }...\overset { \rightarrow }{ { F } } _{ n }\) act on the same body, then the total force \(\left( \overset { \rightarrow }{ { F }_{ net } } \right) \) is equivalent to the vectorial sum of the individual forces
(ii) Their net force provides the acceleration. \(\overset { \rightarrow }{ { F } } _{ net }=\overset { \rightarrow }{ { F } } _{ 1 },\overset { \rightarrow }{ { F } } _{ 2 },\overset { \rightarrow }{ { F } } _{ 3 }...\overset { \rightarrow }{ { F } } _{ n }\)
(iii) Newton's second law for this case is \(\overset { \rightarrow }{ { F } } _{ net }=m\overset { \rightarrow }{ a } \)
In this case the direction of acceleration is in the direction of net force.

(i) Inertia of rest: When a stationary bus starts to move, the passengers experience a sudden backward push.
(ii) Due to inertia, the body (of a passenger) will try to continue in the state of rest, while the bus moves forward. This appears as a backward push.

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
\(0 =m_{1} v_{1} \sin \theta_{1}-m_{2} v_{2} \sin \theta_{2}  \)
\(m_{1} v_{1} \sin \theta_{1} =m_{2} v_{2} \sin \theta_{2}  \)
\(m_{1} =m_{2}  \)
\(\therefore v \sin \theta =\frac{v}{2} \sin 90^{\circ}  \)
\(\sin \theta =\frac{1}{2} \times 1=\frac{1}{2}  \)
\(\therefore \theta =\sin ^{-1}(0.5)=30^{\circ}\)

As per Newton's Third Law of Motion every action has an equal and opposite reaction. As this law is valid on the physical plane in the physical science, it equally holds well but not completely in the metaphysics and in the spiritual science. In the physical plane, the action and reaction are instantaneous; one follows the other without significant time lapse, where as in the metaphysical science there is always a time lag between actions and their reactions. The complete Karmic theory, the theory of actions and reactions, is closely linked and follow the principle of Newton's Third law of equal actions and reactions.
The only difference and additional parameter in the Karmic theory. of actions/ reactions is "inertia or time lapse" of reactions, which has made it more complex and has created confusion in the mind of a common man. Sometimes reaction cannot be easily linked with the action because the reaction may be entirely different from the action executed. Furthermore, actions done on physical plane, mental plane and on intellectual plane bring different reactions in the metaphysical science. It must be experienced and understood by metaphysical science that behind every action there is a divine force. This force acts on us by His grace similar to the Newton's Third Law of physical science as a result of our past actions and it is governed by the inertial forces with time lag, which varies very widely and is beyond the perception of a common human being. Newton's third law is applicable to only human's actions which involves physical force. Third law is not applicable to human's psychological actions or thoughts.

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 =\(\frac { { m }_{ 2 } }{ { \mu }_{ 2 } } -{ m }_{ 1 }\) 

Yes, μ > 1, friction is stronger than normal force

The instantaneous power (P_inst) is defined as the power delivered at an instant (as time interval approaches zero).
\(P_{inst}=\frac{dW}{dt}\)