Taking surface of table as a reference level (zero potential energy)
Potential energy of chain when 1/nth length hanging from the edge \[=\frac{-MgL}{2{{n}^{2}}}\]
Potential energy of chain when it leaves the table \[=-\frac{MgL}{2}\]
Kinetic energy of chain = loss in potential energy
\[\Rightarrow \] \[\frac{1}{2}M{{v}^{2}}=\frac{MgL}{2}-\frac{MgL}{2{{n}^{2}}}\]
\[\Rightarrow \] \[\frac{1}{2}M{{v}^{2}}=\frac{MgL}{2}\left[ 1-\frac{1}{{{n}^{2}}} \right]\]
\[\therefore \] Velocity of chain \[v=\sqrt{gL\left( 1-\frac{1}{{{n}^{2}}} \right)}\] 
The work done in pulling the dy length of the chain on the table.
dW = F(– dy) [As y is decreasing]
i.e. dW = mgy (– dy)
So the work done in pulling the hanging portion on the table.
\[W=-\int_{L/n}^{0}{mgy\,dy}=-mg\,\left[ \frac{{{y}^{2}}}{2} \right]_{L/n}^{0}\]\[=\frac{mg\,{{L}^{2}}}{2{{n}^{2}}}\]
\[\therefore \] \[W=\frac{MgL}{2{{n}^{2}}}\] [As m = M/L]
Alternative method :
If point mass m is pulled through a height h then work done W = mgh
Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at a height of L/(2n) from the lower end and mass of the hanging part of chain \[=\frac{M}{n}\]
So work done to raise the centre of mass of the chain on the table is given by
\[W=\frac{M}{n}\times g\times \frac{L}{2n}\] [As W = mgh]
or \[W=\frac{MgL}{2{{n}^{2}}}\]
For two particles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] separated by a distance r
Gravitational potential energy \[U=-\frac{G\,{{m}_{1}}{{m}_{2}}}{r}\]
(1) If a body of mass m at height h relative to surface of earth then
Gravitational potential energy \[U=\frac{mgh}{1+\frac{h}{R}}\]
Where R = radius of earth, g = acceleration due to gravity at the surface of the earth.
(2) If h << R then above formula reduces to U = mgh.
(3) If V is the gravitational potential at a point, the potential energy of a particle of mass m at that point will be
U = mV
(4) Energy height graph : When a body projected vertically upward from the ground level with some initial velocity then it possess kinetic energy but its initial potential energy is zero.
As the body moves upward its potential energy increases due to increase in height but kinetic energy decreases (due to decrease in velocity). At maximum height its kinetic energy becomes zero and potential energy maximum but through out the complete motion, total energy remains constant as shown in the figure.
i.e. \[\overrightarrow{F}\propto -\overrightarrow{x\,}\]
or \[\overrightarrow{F}=-\,k\,\overrightarrow{x\,}\] ...(i)
where \[k\] is called spring constant.
If \[x=1,\,F=k\] (Numerically)
or \[k=F\]
Hence spring constant is numerically equal to force required to produce unit displacement (compression or extension) in the spring. If required force is more, then spring is said to be more stiff and vice-versa.
Actually k is a measure of the stiffness/softness of the spring.
Dimension : As \[k=\frac{F}{x}\]
\[\therefore \] \[[k]=\frac{[F]}{[x]}=\frac{[ML{{T}^{-2}}]}{L}\]\[=[M{{T}^{-2}}]\]
Units : S.I. unit Newton/metre, C.G.S unit Dyne/cm.
Note :
| Initial state of the spring | Final state of the spring | Initial position \[({{x}_{1}})\] | more...
Potential energy is defined only for conservative forces. In the space occupied by conservative forces every point is associated with certain energy which is called the energy of position or potential energy. Potential energy generally are of three types : Elastic potential energy, Electric potential energy and Gravitational potential energy.
(1) Change in potential energy : Change in potential energy between any two points is defined in the terms of the work done by the associated conservative force in displacing the particle between these two points without any change in kinetic energy.
\[{{U}_{2}}-{{U}_{1}}=-\int_{\,{{r}_{1}}}^{\,{{r}_{2}}}{\vec{F}.\,d\vec{r}=-W}\] ...(i)
We can define a unique value of potential energy only by assigning some arbitrary value to a fixed point called the reference point. Whenever and wherever possible, we take the reference point at infinity and assume potential energy to be zero there, i.e. if we take \[{{r}_{1}}=\infty \] and \[{{r}_{2}}=r\] then from equation (i)
\[U=-\int_{\,\infty }^{\,r}{\vec{F}.\,d\vec{r}=-W}\]
In case of conservative force (field) potential energy is equal to negative of work done by conservative force in shifting the body from reference position to given position.
This is why, in shifting a particle in a conservative field (say gravitational or electric), if the particle moves opposite to the field, work done by the field will be negative and so change in potential energy will be positive i.e. potential energy will increase. When the particle moves in the direction of field, work will be positive and change in potential energy will be negative i.e. potential energy will decrease.
(2) Three dimensional formula for potential energy: For only conservative fields \[\vec{F}\] equals the negative gradient \[(-\vec{\nabla })\] of the potential energy.
So \[\vec{F}=-\vec{\nabla }U\] (\[\vec{\nabla }\] read as Del operator or Nabla operator and \[\vec{\nabla }=\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k}\]) Þ \[\vec{F}=-\left[ \frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k} \right]\]
where,
\[\frac{\partial U}{\partial x}=\]Partial derivative of U w.r.t. x (keeping y and z constant)
\[\frac{\partial U}{\partial y}=\]Partial derivative of U w.r.t. y (keeping x and z constant)
\[\frac{dU}{dz}=\]Partial derivative of U w.r.t. z (keeping x and y constant)
(3) Potential energy curve : A graph plotted between the potential energy of a particle and its displacement from the centre of force is called potential energy curve.
Figure shows a graph of potential energy function U(x) for one dimensional motion.
As we know that negative gradient of the potential energy gives force.
\[-\frac{dU}{dx}=F\]
(4) Nature of force
(i) Attractive force :
On increasing x, if U increases,
\[\frac{dU}{dx}=\text{positive}\], then F is in negative direction
i.e. force is attractive in nature.
In graph this is represented in region BC.
(ii) Repulsive force :
On increasing x, if U decreases,
\[\frac{dU}{dx}=\text{negative}\], then F is in positive direction
i.e. force is repulsive in nature.
In graph this is represented in region AB.
(iii) Zero force :
On increasing x, if more...
If a vehicle moves with some initial velocity and due to some retarding force it stops after covering some distance after some time.
(1) Stopping distance : Let m = Mass of vehicle,
\[\upsilon \] = Velocity, P = Momentum, E = Kinetic energy
F = Stopping force, x = Stopping distance,
t = Stopping time
Then, in this process stopping force does work on the vehicle and destroy the motion.
By the work- energy theorem
\[W=\Delta K=\frac{1}{2}m{{v}^{2}}\]
\[\Rightarrow \] Stopping force (F) x Distance (x) = Kinetic energy (E)
\[\Rightarrow \] Stopping distance (x) \[=\frac{\text{Kinetic}\,\text{energy}\,\text{(}E\text{)}}{\text{Stopping}\,\text{force}\,\text{(}F\text{)}}\]
\[\Rightarrow \] \[x=\frac{m{{v}^{2}}}{2F}\] ...(i)
(2) Stopping time : By the impulse-momentum theorem
\[F\times \Delta t=\Delta P\Rightarrow F\times t=P\]
\[\therefore \] \[t=\frac{P}{F}\]
or \[t=\frac{mv}{F}\] ...(ii)
(3) Comparison of stopping distance and time for two vehicles : Two vehicles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are moving with velocities \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\] respectively. When they are stopped by the same retarding force (F).
The ratio of their stopping distances \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{m}_{1}}v_{1}^{2}}{{{m}_{2}}v_{2}^{2}}\] and the ratio of their stopping time \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}{{v}_{2}}}\]
(i) If vehicles possess same velocities
\[{{\upsilon }_{1}}={{\upsilon }_{2}}\]
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\] ; \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
(ii) If vehicle possess same kinetic momentum
\[{{P}_{1}}={{P}_{2}}\]
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=\left( \frac{P_{1}^{2}}{2{{m}_{1}}} \right)\,\left( \frac{2{{m}_{2}}}{P_{2}^{2}} \right)\,=\frac{{{m}_{2}}}{{{m}_{1}}}\]
\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=1\]
(iii) If vehicle possess same kinetic energy
\[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}=1\]
\[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{\sqrt{2{{m}_{1}}{{E}_{1}}}}{\sqrt{2{{m}_{2}}{{E}_{2}}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\]
Note :
If vehicle is stopped by friction then
Stopping distance \[x=\frac{\frac{1}{2}m{{v}^{2}}}{F}\]\[=\frac{\frac{1}{2}m{{v}^{2}}}{ma}\]\[=\frac{{{v}^{2}}}{2\mu g}\]
\[[\text{As}\,\,a=\mu g]\]
Stopping time \[t=\frac{mv}{F}\]\[=\frac{mv}{m\mu \,g}\]\[=\frac{v}{\mu \,g}\]
The energy possessed by a body by virtue of its motion, is called kinetic energy.
Examples : (i) Flowing water possesses kinetic energy which is used to run the water mills.
(ii) Moving vehicle possesses kinetic energy.
(iii) Moving air (i.e. wind) possesses kinetic energy which is used to run wind mills.
(iv) The hammer possesses kinetic energy which is used to drive the nails in wood.
(v) A bullet fired from the gun has kinetic energy and due to this energy the bullet penetrates into a target.
(1) Expression for kinetic energy :
Let m = mass of the body,
u = Initial velocity of the body (= 0)
F = Force acting on the body,
a = Acceleration of the body,
s = Distance travelled by the body,
\[\upsilon \] = Final velocity of the body
From \[{{v}^{2}}={{u}^{2}}+2as\] Þ \[{{v}^{2}}=0+2as\]
\[\therefore \,s=\frac{{{v}^{2}}}{2a}\]
Since the displacement of the body is in the direction of the applied force, then work done by the force is
\[W=F\times \,s\]\[=ma\times \frac{{{v}^{2}}}{2a}\] \[\Rightarrow \]\[W=\frac{1}{2}m{{v}^{2}}\]
This work done appears as the kinetic energy of the body \[KE=W=\frac{1}{2}m{{v}^{2}}\]
(2) Calculus method : Let a body is initially at rest and force \[\overrightarrow{F}\] is applied on the body to displace it through small displacement \[d\vec{s}\] along its own direction then small work done
\[dW=\overrightarrow{F}.d\overrightarrow{s\,}=F\,ds\]
\[\Rightarrow \] \[dW=m\,a\,ds\] [As F = ma]
\[\Rightarrow \] \[dW=m\frac{dv}{dt}\,ds\] \[\left[ As\,a=\frac{dv}{dt} \right]\]
\[\Rightarrow \] \[dW=mdv.\frac{ds}{dt}\] Þ \[dW=m\,v\,dv\] ...(i)
\[\left[ As\,\,\frac{ds}{dt}=v \right]\]
Therefore work done on the body in order to increase its velocity from zero to \[\upsilon \] is given by
\[W=\int_{0}^{v}{mv\,dv=m\int_{0}^{v}{v\,dv=m\left[ \frac{{{v}^{2}}}{2} \right]_{0}^{v}}}\] \[=\frac{1}{2}m{{v}^{2}}\]
This work done appears as the kinetic energy of the body \[KE=\frac{1}{2}m{{v}^{2}}\].
In vector form \[KE=\frac{1}{2}m\,(\overrightarrow{v\,}.\,\overrightarrow{v\,})\]
As m and \[\overrightarrow{v\,}.\overrightarrow{v\,}\] are always positive, kinetic energy is always positive scalar i.e. kinetic energy can never be negative.
(3) Kinetic energy depends on frame of reference :
The kinetic energy of a person of mass m, sitting in a train moving with speed \[\upsilon \], is zero in the frame of train but \[\frac{1}{2}m{{v}^{2}}\] in the frame of the earth.
(4) Kinetic energy according to relativity : As we know \[E=\frac{1}{2}m{{v}^{2}}\].
But this formula is valid only for (\[\upsilon \] << c) If \[\upsilon \] is comparable to c (speed of light in free space = \[3\times {{10}^{8}}\,m/s\]) then according to Einstein theory of relativity
\[E=\frac{m{{c}^{2}}}{\sqrt{1-({{v}^{2}}/{{c}^{2}})}}-m{{c}^{2}}\]
(5) Work-energy theorem: From equation (i) \[dW=mv\,dv\].
Work done on the body in order to increase its velocity from u to \[\upsilon \] is given by
\[W=\int_{u}^{v}{mv\,dv}\] \[=m\int_{u}^{v}{v\,dv=m\,\left[ \frac{{{v}^{2}}}{2} \right]_{u}^{v}}\]
\[\Rightarrow \] \[W=\frac{1}{2}m[{{v}^{2}}-{{u}^{2}}]\]
Work done = change in kinetic energy
\[W=\Delta E\]
This is work energy theorem, it states that work done by a force acting on a body is equal to the change in more...
The energy of a body is defined as its capacity for doing work.
(1) Since energy of a body is the total quantity of work done, therefore it is a scalar quantity.
(2) Dimension: \[[M{{L}^{2}}{{T}^{-2}}]\] it is same as that of work or torque.
(3) Units : Joule [S.I.], erg [C.G.S.]
Practical units : electron volt (eV), Kilowatt hour (KWh), Calories (cal)
Relation between different units:
1 Joule = \[{{10}^{7}}\] erg
1 eV = \[1.6\times {{10}^{-19}}\] Joule
1 kWh = \[3.6\times {{10}^{6}}\] Joule
1 calorie = \[4.18\, Joule\]
(4) Mass energy equivalence : Einstein's special theory of relativity shows that material particle itself is a form of energy.
The relation between the mass of a particle m and its equivalent energy is given as
\[E=m{{c}^{2}}\] where c = velocity of light in vacuum.
If \[m=1\,amu=1.67\times {{10}^{-27}}\,kg\]
then \[E=931\,MeV=1.5\times {{10}^{-10}}\,Joule\].
If \[m=1kg\] then \[E=9\times {{10}^{16}}\,Joule\]
Examples : (i) Annihilation of matter when an electron \[({{e}^{-}})\] and a positron \[({{e}^{+}})\] combine with each other, they annihilate or destroy each other. The masses of electron and positron are converted into energy. This energy is released in the form of\[\gamma \]-rays.
\[{{e}^{-}}+{{e}^{+}}\to \gamma +\gamma \]
Each \[\gamma \] photon has energy = 0.51 MeV.
Here two \[\gamma \] photons are emitted instead of one \[\gamma \] photon to conserve the linear momentum.
(ii) Pair production : This process is the reverse of annihilation of matter. In this case, a photon \[(\gamma )\] having energy equal to 1.02 MeV interacts with a nucleus and give rise to electron \[({{e}^{-}})\]and positron \[({{e}^{+}})\]. Thus energy is converted into matter.
(iii) Nuclear bomb : When the nucleus is split up due to mass defect (The difference in the mass of nucleons and the nucleus), energy is released in the form of \[\gamma \]-radiations and heat.
(5) Various forms of energy
(i) Mechanical energy (Kinetic and Potential)
(ii) Chemical energy
(iii) Electrical energy
(iv) Magnetic energy
(v) Nuclear energy
(vi) Sound energy
(vii) Light energy
(viii) Heat energy
(6) Transformation of energy : Conversion of energy from one form to another is possible through various devices and processes.
Various devices for energy conversion from one form to another
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