Total energy of an orbiting Satellite and its Binding Energy

Definition:

The total mechanical energy associated with an orbiting satellite around the planet is the sum of kinetic energy (i.e., due to orbital motion) and potential energy (i.e., the gravitational potential energy of the satellite).

Derivation of the total mechanical energy of the orbiting satellite around the planet:

Let us consider

The mass of the satellite = $m$

The mass of the planet = $M$

The distance between satellite to a planet from the center of the planet = $r$

The radius of planet =$R$

The potential energy of the satellite is

$U=-\frac{G \: M \: m}{r} \qquad(1)$

The kinetic energy of the satellite is

$K=\frac{1}{2}m v_{e}^{2}$

$K=\frac{1}{2} m \left[ \sqrt{\frac{G M}{r}} \right]^{2} \quad \left( \because v_{e}^{2}=\sqrt{\frac{G \: M}{r}} \right)$

$K=\frac{1}{2} \left( \frac{G M m}{r} \right) \qquad(2)$

The total energy (i.e. mechanical energy) of the satellite is

$E= K+U$

Now put the value of kinetic and potential energy from equation $(1)$ and equation $(2)$ in the above equation

$E= \frac{1}{2} \left[ \frac{G M m}{r} \right]+ \left[ -\frac{G M m}{r} \right]$

$E= -\frac{1}{2} \left[ \frac{G M m}{r} \right]$

$E= -\frac{1}{2} \left[ \frac{G M m}{R+h} \right] \left( \because r=R+h \right)$

The above equation shows that the total mechanical energy associated with orbiting satellite is negative.

The total mechanical energy associated with the orbiting satellite around the Earth:

Put $M=M_{e}$ and $R=R_{e}$ then

$E= -\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}+h} \right] \left( \because r=R_{e}+h \right)$

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}+h} \right] \left( \because GM_{e}=g R_{e}^{2} \right)$

If the satellite revolves around near the Earth (i.e., $h=0$) then the total mechanical energy of the satellite

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}+0} \right]$

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}} \right]$

$E= -\frac{1}{2} \left[ g R_{e} m \right]$

Binding Energy of the Satellite:

The minimum amount of mechanical energy required to free the revolving satellite around the planet from its orbit is called the binding energy of the revolving satellite.

We know that, the revolving satellite's total mechanical energy in an orbit of the planet is

$E= -\frac{1}{2} \left[ \frac{G M m}{R+h} \right]$

At infinite distance between a satellite to a planet, the total mechanical energy of the satellite is zero. Therefore, if an orbiting satellite is provided postive energy that is equal to the total mechanical energy, its total mechanical energy becomes zero, and the satellite escapes from the planet's orbit. This total positive mechanical energy is called the gravitational binding energy of the satellite. i.e.

$E= +\frac{1}{2} \left[ \frac{G M m}{R+h} \right]$

The gravitational binding energy associated with orbiting satellite around the earth

$E= +\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}+h} \right]$

If a satellite revolves around near Earth ($h=0$), then the binding energy

$E= +\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}} \right]$

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Expression for Orbital velocity of Satellite and Time Period

Orbital Velocity of Satellite:

When any satellite moves about the planet in a particular orbit then the velocity of the satellite is called the orbital velocity of the satellite.

Expression for Orbital Velocity of Satellite:

Let us consider:

The mass of the satellite = $m$

The mass of planet= $M$

The radius of Planet =$R$

The satellite is moving about the planet at height=$h$

The satellite is moving about the planet with orbital velocity=$v_{\circ}$

The distance from the center of the planet to satellite=$r$

The force of gravitation between the planet and the satellite

$F=G \frac{M \: m}{r^{2}} \qquad(1)$

This force work act as a centripetal force to revolve the satellite around the planet i.e.

$F=\frac{m v_{\circ}^{2}}{r} \qquad(2)$

From the above equation $(1)$ and equation $(2)$, we get

$\frac{m v_{\circ}^{2}}{r} = G \frac{M \: m}{r^{2}}$

$v_{\circ}= \sqrt{\frac{G \: M }{r}}$

Where $r=R+h$ then

$v_{\circ}= \sqrt{\frac{G \: M }{R+h}} \qquad(3)$

This is the equation of the orbital velocity of the satellite.

If any satellite revolves around the earth then the orbital velocity

$v_{\circ}= \sqrt{\frac{G \: M_{e} }{R_{e}+h}}$

$v_{\circ}= \sqrt{\frac{gR_{e}^{2}}{R_{e}+h}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+h}} $

This is the equation of the orbital velocity of a satellite revolving around the earth.

If the satellite is orbiting very close to the surface of the earth ( i.e $h=0$) then the orbital velocity

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+0}} $

$v_{\circ}= \sqrt{g R_{e}} $

Now subtitute the value of radius of earth (i.e $R_{e}=6.4 \times 10^{6} \: m$) and gravitational accelertaion ($g=9.8 \:m/sec^{2}$) then orbital velocity

$v_{\circ}=7.92 \: Km/sec$

The time period of Revolving Satellite:

The time taken by satellite to complete on revolution around the planet is called the time period of the satellite.

Let us consider the time period of the revolving satellite is $T$ Then

$T= \frac{Distance \: covered \: by \: Satellite \: in \: one \: revolution}{Orbital \: Velocity}$

$T= \frac{2 \pi r}{v_{\circ}}$

$T= \frac{2 \pi \left( R+h \right)}{v_{\circ}}$

$T= \frac{2 \pi \left( R+h \right)}{v_{\circ}}$

Now subtitute the value of orbital velocity $v_{\circ}$ from equation $(3)$ in above equation then

$T= \frac{2 \pi \left( R+h \right)}{\sqrt{\frac{G \: M }{R+h}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R+h \right)^{3}}{G M}}$

This is the equation of the time period of the revolution of satellites.

If a satellite revolves around the earth then the time period

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{G M_{e}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{g R_{e}^{2}}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

This is the equation of the time period of revolution of satellite revolving around the earth.

If the satellite revolves very nearly around the earth (i.e $h=0$) then the time period of the satellite from the above equation

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+0 \right)^{3}}{g R_{e}^{2}}} $

$T= 2 \pi \sqrt{ \frac{ R_{e}}{g}} $

$T= 84.6 \: min $

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Derivation of Gravitational Potential Energy due to Point mass and on the Earth

Definition of Gravitational Potential Energy:

When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is stored in the form of potential energy which is called gravitational potential energy.

Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential energy

$U=W_{\infty \rightarrow r}$

Derivation of Gravitational Potential energy due to a Point mass:

Let us consider,
The mass of the point object (i.e point mass)=$m$
The mass of the object that produces the gravitational field = $M$

If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is

$F=G \frac{M \: m}{x^{2}} \qquad(1)$

If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M \: m}{x^{2}}.dx$

Therefore, the work done to bring the point mass from infinity to point $P$ that is at distance $r$ from point $O$ then work done to move the point object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M\: m }{r} $

This work done by the force is stored in the form of potential energy i.e

$U=W$

$U=-\frac{G\: M\: m }{r}$

Thus the above equation represents the gravitational potential energy of an object at point $P$

Gravitational Potential Energy on Earth:

Let us consider, The mass of Earth = $M_{e}$
The radius of earth = $R_{e}$
The mass of the object = $m$
The distance from centre $O$ of the earth to point $P$ = $r$
The distance from the surface of the earth to point $P$ = $h$

If the object is at a distance $x$ then the gravitational force is

$F=G \frac{M_{e} \: m}{x^{2}} \qquad(1)$

If the object moves a very small distance element $dx$ that is at distance $x$ from centre point $O$ of the earth then the work done to move an object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M_{e} \: m}{x^{2}}.dx$

Therefore, the work done to bring the object from infinity to point $P$ that is at distance $r$ from centre point $O$ of the earth then the work done to move an object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M_{e} \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M_{e}\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M_{e}\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M_{e}\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M_{e}\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M_{e}\: m }{r} $

The above equaton shows that the work done by force is stored in the form of gravitational potential energy i.e.

$U=W$

$ U = -\frac{G\: M_{e}\: m }{r}$

Where $r=R_{e}+h$, then above equation can be written as

$U = -\frac{G\: M_{e}\: m }{R_{e}+h} \qquad(2)$

This is the equation of the gravitational potential energy at point $P$. The other form of the above equation i.e

$U = -\frac{g R_{e}^{2} \: m }{R_{e}+h} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

If the object is placed on the surface of the earth then $h=0$. So gravitational potential energy on the surface of the earth
$U=-\frac{G \: M_{e} \: m}{R_{e}}$

This is the equation of the gravitational potential energy of an object placed on the surface of the earth.

$U=-\frac{g R_{e}^{2} m}{R_{e}} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

$U=-g R_{e} m$ This is another form of the gravitational potential energy of an object placed on the surface of the earth.

Note:

We know that the gravitational potential energy at any point from above the surface of the earth

$U = -\frac{G\: M_{e}\: m }{R_{e}+h} $

$U = V m \qquad \left( \because V= -\frac{G\: M_{e}\: }{R_{e}+h} \right)$

$U = Gravitational \: Potential \times \: mass \: of \: an \: object$

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Derivation of Gravitational Potential due to Point mass and on the Earth

Definition of Gravitational Potential:

When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is called gravitational potential.

Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential

$V=-\frac{W}{m}$

Derivation of Gravitational Potential due to a Point mass:

Let us consider,
The mass of the point object (i.e point mass)=$m$
The mass of the object that produces the gravitational field = $M$

If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is

$F=G \frac{M \: m}{x^{2}} \qquad(1)$

If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M \: m}{x^{2}}.dx$

Therefore, the work done to bring the point mass from infinity to point $P$ that is at distance $r$ from point $O$ then work done to move the point object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M\: m }{r} $

$\frac{W}{m} = -\frac{G\: M }{r} $

$ V = -\frac{G\: M}{r} \qquad \left( \because V=\frac{W}{m} \right)$

$ V = -\frac{G\: M}{r}$

Thus the above equation represents the gravitational potential at point $P$

Gravitational Potential on Earth:

Let us consider, The mass of Earth = $M_{e}$
The radius of earth = $R_{e}$
The mass of the object = $m$
The distance from centre $O$ of the earth to point $P$ = $r$
The distance from the surface of the earth to point $P$ = $h$

If the object is at a distance $x$ then the gravitational force is

$F=G \frac{M_{e} \: m}{x^{2}} \qquad(1)$

If the object moves a very small distance element $dx$ that is at distance $x$ from centre point $O$ of the earth then the work done to move an object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M_{e} \: m}{x^{2}}.dx$

Therefore, the work done to bring the object from infinity to point $P$ that is at distance $r$ from centre point $O$ of the earth then the work done to move an object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M_{e} \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M_{e}\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M_{e}\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M_{e}\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M_{e}\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M_{e}\: m }{r} $

$\frac{W}{m} = -\frac{G\: M_{e} }{r} $

$ V = -\frac{G\: M_{e}}{r} \qquad \left( \because V=\frac{W}{m} \right)$

$ V = -\frac{G\: M_{e}}{r}$

Where $r=R_{e}+h$, then above equation can be written as

$V=-\frac{G \: M_{e}}{R_{e}+h} \qquad(2)$

This is the equation of gravitational potential at point $P$. The other form of the above equation i.e

$V=-\frac{g R_{e}^{2}}{R_{e}+h} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

If the object is placed on the surface of the earth then $h=0$. So gravitational potential on the surface of the earth

$V=-\frac{G \: M_{e}}{R_{e}}$

This is the equation of gravitational potential on the surface of the earth. The other form of the above equation is

$V=-\frac{g R_{e}^{2}}{R_{e}} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

$V=-g R_{e}$

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Gravitational field, Intensity of Gravitational field and its expression

Definition of Gravitational Field:

The region around an object in which another object experiences a gravitational force then the region of that object is called the gravitational field.

Gravitational Field Intensity:

The force applied per unit mass of an object that is placed in the gravitational field is called the intensity of the gravitational field.

$\overrightarrow{E}=- \frac{G \: M}{r^{2}} \hat{r}$

The Expression for gravitational field intensity:

Let us consider
The mass of a lighter object that experience the force = $m$
The mass of a heavy object that produces the gravitational field= $M$
The distance between the objects = $r$

The gravitational force between the objects is

$F=G\frac{M\:m}{r^{2}} \qquad(1)$

Now the force per unit mass i.e Gravitational field intensity

$E=-\frac{F}{m} \qquad (2)$

Here the negative indicates that the direction of force is opposite to $\hat{r}$

The vector form of the gravitational field intensity

$\overrightarrow{E}=-\frac{F}{m} \hat{r}$

Where $\hat{r} \left (=\frac{\overrightarrow{r}}{r} \right)$ is the unit vector along the $\overrightarrow{r}$

Now substitute the value of $F$ in the above equation $(2)$. Therefore we get,

$E=-G \frac{M\:m}{m r^{2}}$

$E=- \frac{G \: M}{r^{2}}$

The vector form of the above equation

$\overrightarrow{E}=- \frac{G \: M}{r^{2}} \hat{r}$

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Deduction of Newton's Law of gravitation from Kepler's Law

Deduction of Newton's Law of gravitational force from Kepler's Law:

Let us consider:
The mass of a planet = $m$
The radius of the circular path of a planet=$r$
The mass of the sun = $M$
The velocity of the planet = $v$
The time of the revolution=$T$

The attraction force between the planet and the sun is achieved by the centripetal force i.e

$F=\frac{m v^{2}}{r} \qquad(1)$

The orbital velocity of the planet:

$v=\frac{Circumference \: of \: the \: circular \: path}{Time \: period}$

$v=\frac{2\pi r}{T} \qquad(2)$

From equation $(1)$ and equation $(2)$, we get

$F=\frac{m}{r} \left( \frac{2\pi r}{T} \right)^{2}$

$F=\frac{4 \pi^{2} mr}{T^{2}} \qquad(3)$

According to Kepler's third law i.e

$T^{2}=Kr^{3} \qquad(4)$

From equation$(3)$ and equation$(4)$, we get

$F=\frac{4 \pi^{2}mr}{Kr^{3}}$

$F=\frac{4 \pi^{2}}{K}\frac{mr}{r^{3}}$

$F=\frac{4 \pi^{2}}{K}\frac{m}{r^{2}} \qquad(5)$

The source of this force is the sun. So R.H.S of equation $(5)$ should be related to the sun. Since $m$ and $r$ are related to the planet. So the quantity $\frac{4 \pi^{2}}{K}$ should be related to some constant of the sun. Let $\frac{4 \pi^{2}}{K}$ be proportional to the mass of the sun. i.e

$\frac{4 \pi^{2}}{K m} \propto M$

$\frac{4 \pi^{2}}{K m} = G M $

Here $G$ is the proportionality constant and it is called the universal gravitation constant. Now substitute the $\frac{4 \pi^{2}}{K m} = G M$ in equation $(5)$. then we get

$F=G \frac{M m}{r^{2}}$

The above equation represents the force of attraction between the sun and the planet. It is also known as Newton's law of gravitational force.

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Force of Friction | Expression for the acceleration and the work done on inclined rough surface| Advantage and Disadvantage

Definition of Friction:

When an object slides over the surface of another object then each surface applies a parallel force to the other. This force is called friction force. The direction of the force of friction on every object with respect to the other is always opposite to the motion of the second object.

$F=\mu R$

Properties of Friction Force:

The force of friction always acts in the opposite direction of the motion of the object.

The force of friction always opposes the motion of the object. So it does not help in motion.

Friction force always helps an object to be in the rest of the position. i.e If an object is in the rest of the position that is due to only frictional force.

The frictional force is act on both states of the object i.e Rest and motion.

Modern Concept of Friction:

It is now believed that frictional forces arise because of a fundamental force occurring in nature commonly known as an electric force. Every object has surface irregularities at the atomic level.

When two surfaces come in contact then the real area in contact is much smaller than the actual area of the surface. At the contact points, the distance between the particles becomes very small and as such the molecular electrical forces begin to act between the surfaces and molecular bonds are formed.

When one of the surfaces is pulled over the other, the molecular bonds are broken. As a result, the material of the objects is deformed and new bonds are formed. The successive loading and deformation processes result in loss of energy which appears in the form of heat. To compensate for the energy loss, a force is required to be applied to start the motion.

Types of forces of friction: There are two types of forces of friction

1. Force of Static Friction
2. Force of Dynamic Friction

1. Force of Static Friction:

When the friction forces acting between two surfaces at rest with respect to each other are called the force of static friction.

Note: The maximum value of the force of static friction is always equal to the minimum value of force required to start the motion. Once the motion starts, the force of friction after reducing becomes such that it is sufficient to maintain uniform motion.

Due to the force of static friction between two dry and unlubricator surfaces, the maximum force is according to the following two laws:

i.) The maximum force does not normally depend upon the area of contact.
ii.) It is directly proportional to the normal reaction.

Coefficient of Static Friction:

The ratio of the maximum value of force in static friction and normal reaction is called the coefficient of static friction.

$\mu_{s}=\frac{F}{R}$

2. Force of Dynamic Friction:

When the force of friction acting between the two surfaces in relative motion is called the force of dynamic friction.

Note: Due to dynamic friction between the dry and unlubricated surfaces the value of dynamic friction is according to the two laws:

i.) It does not depend upon the surfaces of contact.
ii.) It is equal to the maximum normal reaction force.

Coefficient of Dynamic Friction:

The ratio of force in dynamic friction and normal reaction is called the coefficient of dynamic friction.

$\mu_{k}=\frac{F}{R}$

Limiting friction:

It is the maximum value of friction that acts when the object just begins to move is called Limiting Friction. The limiting friction is directly proportional to the normal reaction in the object. i.e

$F \propto R$

$F = \mu R$

Angle of friction:

The angle between the effective resistance $S$ (i.e resultant of frictional force and normal reaction) and normal reaction $R$ is called the angle of friction.

Let us consider, A object of mass $m$ placed on a horizontal surface. If $F$ force is applied on the object then the value of limiting friction

$F \propto R$

$F = \mu R$

$\mu=\frac{F}{R} \qquad(1.1)$

From figure in $\Delta ROS$

$tan \theta =\frac{RS}{RO}$

$tan \theta =\frac{RS}{RO}$

$tan \theta =\frac{F}{R} \qquad \left( \because RS=F \: and \: RO=R \right)$

From the equation $(1.1)$ and the above equation, We can conclude that

$\mu=tan\theta$

Angle of Repose and Angle of Sliding:

The angle of repose or angle of sliding is defined as the minimum angle of inclination of a surface with the horizontal, such that an object placed on the surface just begins to slide down.

The various forces acting on the object are shown in the figure below:

i.) Weight $mg$ of the object acting vertically downwards.
ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
iii.) Force of friction $f$ acting up in the inclined surface $AB$

Now from the figure apply the equilibrium condition:

$R=mg \: cos\theta \qquad(1.2)$

$f=mg \: sin\theta \qquad(1.3)$

Now divide by equation $(1.3)$ to equation $(1.2)$

$\frac{f}{R}=\frac{mg \: sin\theta}{mg \: cos\theta}$

$\frac{f}{R}=\frac{sin\theta}{cos\theta}$

$\frac{f}{R}=tan \theta$

$\frac{\mu R}{R}=tan \theta \qquad \left( \because f=\mu R \right)$

$\mu =tan \theta $

Thus the angle of friction is equal to the angle of repose.

1.) Expression for the acceleration and work done on the object moves downward on a rough inclined surface without applied force:

Let us consider, A object of mass $m$ moving sliding downward on a rough surface that is inclined at an angle $\theta$ from horizontal. If the angle $\theta$ of the inclined surface is greater than the angle of repose, the object slides down with an acceleration $a$ without any applied force.

The various forces acting on the object are shown in the figure below:

i.) Weight $mg$ of the object acting vertically downwards.
ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
iii.) Force of friction $f$ acting up in the inclined surface $AB$

Apply the equilibrium condition from the figure:

$R=mg \: cos\theta \qquad(1.4)$

The net force on the object moving down the inclined surface

$F=mg \: sin\theta - f \qquad(1.5)$

$F=mg \: sin\theta - \mu R $

Now substitute the value of $R$ from equation $(1.4)$ in above equation

$F=mg \: sin\theta - \mu mg \: sin \theta$

$ma=mg \left( \: sin\theta - \mu \: sin \theta \right) \qquad( F=ma)$

$a=g \left( \: sin\theta - \mu \: sin \theta \right) $

If the object is displaced the distance $S$ by the acceleration $a$ then work done by force:

$W=F.S$

$W= mg \left(\: sin\theta - \mu \: sin \theta \right). S$

2.) Expression for acceleration and work done on the object moving upward on a rough inclined surface by the applied force:

Let us consider, A object of mass $m$ moving upward on a rough inclined surface by applying the force $F$ sliding downward on a rough surface so the various forces acting on the object are shown in the figure below:

i.) Weight $mg$ of the object acting vertically downwards.
ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
iii.) Force of friction $f$ acting up in the inclined surface $AB$

In equilibrium condition from the figure:

$R=mg \: cos\theta \qquad(1.6)$

The net force on the object moving down the inclined surface

$F=mg \: sin\theta + f \qquad(1.7)$

$F=mg \: sin\theta + \mu R $

Now substitute the value of $R$ from equation $(1.6)$ in above equation

$F=mg \: sin\theta + \mu mg \: sin \theta$

$ma=mg \left( \: sin\theta + \mu \: sin \theta \right) \qquad( \because F=ma)$

$a=g \left( \: sin\theta + \mu \: sin \theta \right) $

If the object is displaced the distance $S$ by the force $F$ then work done by force:

$W=F.S$

$W= mg \left(\: sin\theta + \mu \: sin \theta \right). S$

3.) Expression for acceleration and work done on the object moves on a rough horizontal surface by the applied force:

Let us consider, A object is moving on a rough horizontal surface by applying the force $F$ on the object of mass $m$ and then the net force on the object from the figure below

$F-f=ma$

$F-\mu R=ma \qquad \left(\because f=\mu R \right)$

$F-\mu mg=ma \qquad \left(R=mg \right)$

$F=m \left(\mu g+a \right)$

$a=\frac{F-\mu mg}{m}$

If the object is displaced the distance $S$ by the force $F$ then work done by force:

$W=F.S$

$W= m \left(\mu g+a \right). S$

The Advantages of Friction:

The two objects will not stick to each other if there is no friction between the surface.

The parts of machinery are held together with the help of nuts and bolts but without friction, these can not be held.

A person can not walk or stand on the surface without friction.

The brakes of any vehicle will not work without friction.

It is not possible to transfer motion from one part of a machine to the other part without the help of friction.

Sandpaper is used in cleaning because this cleaning is only possible with the help of friction.

Adhesives will lose their purpose.

The Disadvantage of Friction:

Friction always opposes the relative motion between any two objects in contact. Therefore, extra energy is lost to overcome friction. Thus, friction involves the unnecessary loss of energy. This shows that the output is always less than the input.

Friction between the parts of machinery causes wears and tear. Therefore, the lifetime of the parts of the machinery reduces.

Frictional forces produce heat, which causes damage to the machinery.

Methods of Reducing friction:

By Polishing: Polishing makes the surface smoother. Therefore, friction reduces.

By Streamlining: Friction due t air is considerably reduced by streamlining the shape of the object (sharp in front) moving through the air.

By Lubrication: Oil, grease, and many other materials are used as lubricants. These lubricants fill up the irregularities of the surface making them smother. Hence, friction decreases.

By proper selection of materials.

By using ball bearings.

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