Work power energy is a very important concept in physics.work done by all the forces is equal to the change in kinetic energy.
Work : work is related to the amount of energy transferred to or from a system by a force.
The work done by a constant force F acting parallel to the displacement \(\vartriangle \) x is defined as Wf = F \(\vartriangle \) x.
When the force makes a body move,then work is said to be done.But for doing work,energy is required. “Power is the work done in a unit of time.In other words,power is a measure of how qickly work can be done
The unit of power us the watt = 1 joule /1sec.
One common unit of energy is the kilowatt-hour(KWh).”
Exp: “ lifting a weight from the ground and putting it on a shelf is a good example of work.the force is equal to tha weight of the object,and the distance is equal to the height of the shelf(W=FXd).
Workdone:
Work done is defined as product of the force and the distance over which the force is applied.work is done when a force is applied to an object and the object is moved through a distance.
For Exp: when you lift a load you are applying a force over a distance so you are doing work.
“workdone and energy transferred are measured in joles(J).”
Formula:
Workdone by driving forc = change in kinetic energy(K.E)+ change in potential enrgy(P.E)-workdone against resistance
Positive Work:
Work done is said to positive if applied force or one of its component is in the direction of displacement when \(0{}^\circ \le \theta <90{}^\circ \)
Therefore \(W=\left( FS\cos \theta \right)\) is positive
Ex 1 : When a horse pulls a cart the applied force and displacement are in the same direction so work done by horse is positive.
Ex 2: Work done by the gravitational force on a freely falling body is positive
Ex 3: When a spring is stretched, both the stretching force and displacement act in the same direction so work done is positive.
Ex 4: When a block is lifted from the ground the work done by the lifting force is positive
Negative Work:
Work done by a force is said to be negative if the applied force has a component in a direction opposite to that of the displacement when \(90{}^\circ <\theta \le 180{}^\circ \) then \(\cos \theta \) is negative therefore \(W=\left( FS\cos \theta \right)\) is negative.
Ex 1 : When a body is dragged on rough surface, work done by frictional force is negative
Ex 2: Work done by the gravitational force on a vertically projected up body is negative.
Zero Work:
Ø If a body displaces perpendicular to the direction of force then work done is zero \(\left( \theta ={{90}^{O}} \right)\)
Ø If there is no displacement [S=0] then work done is zero
Ø If the applied force is zero(F=0), then work done is zero
Examples:
Ø A person carrying a load and moving horizontally on a platform does no work against gravity.
Ø When a body moves in a circle the work done by the centripetal force is zero.
Ø The tension in the string of a simple pendulum is always perpendicular to its displacement, So work done by tension is zero.
Ø When a person tries to displace a wall by applying a force and if it does not move the work done by him is zero.
Ø A person carrying a load on his head and standing at a given place does no
1) In which of the following, the work done by the mentioned force is negative? The work done by
Soul: the gravitational force when a body is thrown up.
2) Work done by force of friction.
Ans: can be zero(or) be positive (or)be negative.
The work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product: W = Fd. Integration approach can be used both to calculate work done by a variable force and work done by a constant force.
When the magnitude and direction of a force varies with position, the work done by such a force for an infinitesimal displacement ds is given by \(dw=\bar{F}.\overline{ds}\)
The total work done in going from A to B is \({{W}_{AB}}=\int\limits_{A}^{B}{\overrightarrow{F}.\overrightarrow{ds}}=\int\limits_{A}^{B}{\left( F\cos \theta \right)ds}\)
In terms of rectangular components
\(\vec{F}={{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}+{{F}_{z}}\hat{k}\)
\(\overrightarrow{ds}=dx\hat{i}+dy\hat{j}+dz\hat{k}\)
\(W=\int\limits_{A}^{B}{{{F}_{x}}}dx+\int\limits_{A}^{B}{{{F}_{y}}}dy+\int\limits_{A}^{B}{{{F}_{z}}}dz\)
Kinetic energy of a particle moving in a straight line varies with time t as \(K=4{{t}^{2}} \).The force acting on the particle
\(\frac{1}{2}m{{v}^{2}}=4{{t}^{2}}\) \(\therefore v=\sqrt{\frac{8}{m}}t\)
comparing with v=at, a = constant
i.e.,the force acting on the particle is constant
Ans: 30 \(\sqrt{3}\) J
2.A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = -a to x=+2a
Ans: \(\frac{3ab}{2}\)
Graphical Interpretation of Work done:
Area of F-S graph gives work
Ø Work done by constant force.
The area enclosed by the graph on displacement axis gives the amount of work done by the force
Work = FS = Area of OPQR
Ø Work done by variable force
For a small displacement dx the work done will be the area of the strip of width dx
\(W=\int\limits_{{{X}_{i}}}^{{{X}_{f}}}{dw}=\int\limits_{{{X}_{i}}}^{{{X}_{f}}}{F\ dx}\)
In this case work done is positive
If area lies above X-axis work done is +ve if the area lies below X-axis work done is –ve
Illustrations:
1)A particle moves under the effect of a force F = C x from x = 0 to x = x1. The work done in the process is (treat C as a constant)
Ans:
½ Cx12
2) The work done by a force in displacing a particle from x=4m to x=-2m is
Ans: 360J
Ø If force is changing linearly from to over a displacement S then work done is
\(W=\left( \frac{{{F}_{1}}+{{F}_{2}}}{2} \right)s\)
Ø If force displaces the particle from its initial position to final position then displacement vector
\(\overrightarrow{S}=\overrightarrow{{{r}_{f}}}-\overrightarrow{{{r}_{i}}}\)
\(W=\overrightarrow{F}.\overrightarrow{S}=\overrightarrow{F}.\left( \overrightarrow{{{r}_{f}}}-\overrightarrow{{{r}_{i}}} \right)\)
Ø An object of mass m is lifted to a height h above the ground without any acceleration.
Work done by lifting force
\({{W}_{F}}=Fh=mgh\)
Work done by the gravitational force is
\(Wg=-mgh\)
CONCEPTUAL UNDERSTANDING:
1) Two springs have their force constants k1 and k2 (K2>K1). When they are stretched by the same force, work done is
Ans: more in spring K1..
Ø Energy is ability or capacity to do work. Greater the amount of energy possessed by the body, greater the work it will be able to do.
Ø Energy is cause for doing work and work is effect.
Ø Energy is a scalar. Energy has units and dimensions as of work.
i.e ”energy in the process of transfer from one body to another..for example,any given body has kinetic energy if it is in motion.”
Ø The different forms of energy are Mechanical energy, Light Energy, Heat Energy, Sound energy, Electrical energy, Nuclear energy etc.
Mechanical energy:
Mechanical energy is the sum of potential energy and kinetic energy.it is the energy associated with the motion and position of an object.
Mechanical energy is of two type
a) Kinetic Energy
b) Potentical Energy
Kinetic energy is the energy possessed by a body by virtue of its motion.
Kinetic Energy of abody of mass ‘m’ moving with a velocity ‘v’ is \(KE=\frac{1}{2}m{{v}^{2}}\)
since m and v are positive KE is always positive. Kinetic energy is a scalar quantity.
The kinetic energy of an object is a measure of the work an object can do by the virtue of its motion
Examples for bodies having K.E:
1) A vehicle in motion
2) Water flowing along a river
3) A bullet fired from a gun
Expression for KE :
Consider a body of mass ‘m’ moving with a velocity ‘’. A uniform force opposes its motion to bring this body to rest in a displacement s. The uniform retardation of the body due to the force is ‘a’ which is obtained by the equation.
\({{\nu }^{2}}-{{u}^{2}}=2as\Rightarrow o-{{v}^{2}}=2as\)
\(a=-\frac{{{v}^{2}}}{2s}\)
The opposing force on the body is
\(F=ma=\frac{m{{v}^{2}}}{2s}\)
From Newton’s third law of motion. Force applied by the body \(=-F=\frac{m{{v}^{2}}}{2s}\)
As the body moves against the applied force, its displacement is along the force applied by the body here workdone by the body against the opposing force acting on it is
\(w=\left( \frac{m{{v}^{2}}}{2s} \right)\left( s \right)\cos {{0}^{0}}=\frac{1}{2}m{{v}^{2}}\)
Kinetic energy of the body \(KE=\frac{1}{2}m{{v}^{2}}\)
Ø When a body of mass m moves with a velocity ‘v’ its kinetic energy is \(KE=\frac{1}{2}m{{v}^{2}}\) If two bodies of different
Masses have same velocity then \(KE\,\alpha \,m\)
Øalso \(KE=\frac{1}{2}m{{v}^{2}}=\frac{{{m}^{2}}{{v}^{2}}}{2m}=\frac{{{\left( mv \right)}^{2}}}{2m}=\frac{{{p}^{2}}}{2m}\)
Ø If the two bodies of different masses have same momentum then lighter body will have greater K.E \(\left( \because KE\alpha \frac{1}{m} \right)\)
Ø When a bullet is fired from a gun the momentum of the bullet and gun are equal and opposite. It can be seen from the above expression that the ratio of kinetic energics of two bodies having the same magnitude of momenta is in the inverse ratio of their masses
Hence, the KE of the bullet is greater than that of the gun
Ø A body can have energy without momentum. But it can not have momentum without energy.
Ø A bullet of mass ‘m’ moving with velocity ‘v’ stops in wooden block aftger penterating through a distance x. If F is resistance offered by the block of the bullet (Assuming F is Constant inside the block)
\(\frac{1}{2}m{{v}^{2}}=Fx\)
\(F=\frac{m{{v}^{2}}}{2x}\)
1) The change in kinetic energy per unit ‘space’ (distance) is equal to
Ans: force
2 .When the momentum of a body is doubled, the kinetic energy is
Ans: becomes four times
3. Two bodies of masses m1 and m2 have equal momentum. Their K.E. are in the ratio
Ans: m2:m1
Ans :
K/4
Ans :
3 J
Potential energy of a body is the enrgy possessed by a body by virtue of its position or configuration in the gravitational field.
Potential energy is defined only for conservative forces. It does not exist for non-conservative forces. In case of conservative forces.
\(F=-\left( \frac{dU}{dr} \right)\)U is P.E
\(\therefore dU=-\bar{F}.\overline{dr}\)
\(\int\limits_{{{u}_{1}}}^{{{u}_{2}}}{dU}=-\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\overline{dr}}\)
\({{U}_{2}}-{{U}_{1}}=-\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\,d\overline{r}}=-w\)
\(\left( \because \int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\,d\overline{r}=w} \right)\)
\(\therefore \,\,\,If\,\,{{r}_{1}}=\infty \,,\,{{U}_{1}}=0\)
\(\therefore \,\,U=-\int\limits_{\infty }^{r}{\overline{F}.d\overline{r}=-w}\)
P.E can be +ve or -ve or can be zero.
P.E depends on frame of reference.
Examples: water stored in a dam possesses P.E
A streteched bow possesses P.E
Gravitational PE:
A body of mass m is at a height h above the ground. Its PE \(U=\frac{mgh}{1+\frac{h}{R}}\)
where R is radius of the earth, g-acceteration due to gravity.
If h <<< R then , \(\frac{h}{R}<<<1\) we can neglect \(\frac{h}{R}\)
U= mgh
A moving body may or may not have P.E
If P.E of the system is negative then the particles are bounded to the system.
If PE of the system is positive the particles are not bounded to the system.
Ø The absoulte value of PE has no physical significance.
Spring force is an example of a variable force which is conservative. The figure shows one end of a massless spring attached to a rigid vertical support and the other end to a block of mass ‘m’ which can move on a smooth horizontal surface.
Let x = 0 denote the position of block when the spring is at its natural length. In an ideal spring, the spring force is directly proportional to ‘x’
where x is the displacement of the block from equilibrium position. This can be expressed as \({{F}_{s}}=-Kx\)
The constant K is clled the spring constant.
To calculate the workdone on the block by the spring force as the block moves from undeformed position
\(x=0\,to\,x={{x}_{1}}\)
\(dw=\overline{F}.d\overline{x}=-kxdx\)
\(\int{dw=\int\limits_{0}^{{{x}_{1}}}{-kxdx}}\)
\(\therefore \,\,{{W}_{1}}=-\frac{1}{2}k\left( {{x}^{2}} \right)_{0}^{{{x}_{1}}}=-\frac{1}{2}kx_{1}^{2}\)
The workdone by the spring force is always negative. similar is the result of the workdone by spring force when it is compressed through from its natural length \(w=-\frac{1}{2}kx_{1}^{2}\)
If the block moves from \(x={{x}_{1}}\,to\,x={{x}_{2}}\)
workdone by spring force is \(w=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{-kxdx}\)
\(w=\,\frac{1}{2}k\left( x_{1}^{2}-x_{2}^{2} \right)=\frac{1}{2}kx_{1}^{2}-\frac{1}{2}kx_{2}^{2}\)
The change in potential energy of a system corresponding to a conservative internal force is
\(dU=-\int\limits_{0}^{x}{\overline{F}\,\overline{dx}}\)
dU = - (work done by the spring forces)
\(dU=-\left( \frac{-k{{x}^{2}}}{2} \right)\)
\({{U}_{f}}-{{U}_{i}}=\frac{1}{2}k{{x}^{2}}\)
since is zero when spring is at its natural length
\(\therefore {{U}_{f}}=\frac{1}{2}k{{x}^{2}}\)
Electric Potential Energy:
Consider two point charges \({{q}_{1}},{{q}_{2}}\),\({{q}_{1}},{{q}_{2}}\) seperated by a distance r. Then the PE of two point
charges \(PE=\frac{1}{4\pi {{\in }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{r}\) where is permitivity of free space.
Conceptual understanding:
1) If X, F and U denote the displacement, the force acting on and potential energy of a particle, then
Ans: \(F=-\frac{dU}{dx}\)
2) When a spring is wound, a certain amount of PE is stored in it. If this wound spring is dissolved in acid the stored energy
Ans: Appears in the form of heat raising the temperature of the acid.
Ans : 5 107 J
Ans : 8.82J
Statement: Work done by all forces acting on a body is equal to change in its kinetic energy.
Proof:
Consider a particle of mass ‘m’ moving with an initial velocity ‘u’. When it is under the action of a constant net force F, let it gain uniform acceleration ‘a’. Its velocity becomes ‘v’ after a displacement s.
Work done by the net force W=FS (F=ma)
W=mas \(\left( \because a=\frac{{{v}^{2}}-{{u}^{2}}}{2s} \right)\)
\(w=m\left( \frac{{{v}^{2}}-{{u}^{2}}}{2} \right)\)
\(w=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}\)
\(\therefore W={{K}_{f}}-{{K}_{i}}\)
\(W=\Delta K\)
Where Kf and Ki are the final and initial kinetic energies of the particles. Kf - Ki is the change in kinetic energy of the particle.
Ø Work energy theorem is applicable not only for a single particle but also for a system.
When it is applied to a system of two or more particles change in the kinetic energy of the system is equal to workdone on the system by the external as well as the internal forces.
Ø Work-energy theorem can also be applied to a system under the action of variable forces, conservative as well as non-conservative forces.
Ø A body of mass m starting from rest acquired a velocity ‘v’ due to constant force F. Neglecting air resistance.
Work done = Kinetic energy = KE
\(W=\frac{1}{2}m{{v}^{2}}\)
Ø A particle of mass ‘m’ is thrown vertically up with a speed ‘u’. Neglecting the air friction, the work done by gravitational force, as particle reaches maximum height is
\({{W}_{g}}=\Delta K={{K}_{f}}-{{K}_{i}}\)
\({{W}_{g}}=-\frac{1}{2}m{{u}^{2}}\ \ \left( \because {{K}_{f}}=\frac{1}{2}m\times 0 \right)\)
Ø A particle of mass ‘m’ falls freely from a height ‘h’ in air medium onto the ground. If ‘u’ is the velocity with which it reaches the ground, the work done by air friction is Wf and work done by gravitational force (Wg = mgh)
\(\because {{W}_{g}}+{{W}_{f}}=\Delta K\)
Ø A block of mass ‘m’ slides down a frictionless smooth incline of inclination ‘ ’o the horizontal.If h is the height of incline, the velocity with which the body reaches the bottom of incline is
\({{w}_{g}}=\Delta K\)
\(mgh=\frac{1}{2}m{{v}^{2}}-0\)
\(mgh=\frac{1}{2}m{{v}^{2}}\)
\(v=\sqrt{2gh}\)
Ø A body of mass ‘m’ starts form rest from the top of a rough inclined plane of inclination \(\theta \) and length ‘l’. The velocity ‘v’ with which it reaches the bottom of incline if is the coefficient of kinetic friction is
\({{W}_{g}}+{{W}_{f}}=\Delta k\)
\(\left( mg\sin \theta \right)l+\left( -{{\mu }_{k}}mg\cos \theta \right)l=\frac{1}{2}m{{v}^{2}}-0\)
\(v=\sqrt{2gl\left( \sin \theta -{{\mu }_{k}}\cos \theta \right)}\)
Ø Consider a simple pendulum of length l. If it is given a speed u at its lowest position then the speed of the bob when it makes an angle with the vertical is
\({{W}_{g}}=\Delta K\)
\(-mgl\left( 1-\cos \theta \right)=\frac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)\)
\(v=\sqrt{{{u}^{2}}-2gl\left( 1-\cos \theta \right)}\)
Ø A bullet of mass ‘m’ moving with velocity ‘v’ stops in a wooden block after penetrating through a distance x. If ‘F’ is the resistance offered by the block to the bullet.
\({{W}_{F}}={{K}_{f}}-{{K}_{i}}\)
\(-fx=0-\frac{1}{2}m{{v}^{2}}\)
\(f=\frac{1}{2}\frac{m{{v}^{2}}}{x}\)
Let the mass be oscillating with amplitude x on compressing the spring
\({{W}_{S.F}}=-\frac{1}{2}K{{x}^{2}}\)
\({{W}_{g}}=\,mg\,\cos {{90}^{0}}=0\)
\({{W}_{N}}=\,N\,\cos {{90}^{0}}=0\)
Since no work is done by gravity and normal reaction N
\({{W}_{SF}}=\,{{K}_{f}}-{{K}_{i}}\,\)
\(-\frac{1}{2}K{{x}^{2}}=0-\frac{1}{2}m{{v}^{2}}\)
\(x=v\sqrt{\frac{m}{k}}\)
Ø A pile drive of mass ‘m’ is dropped from a height ‘h’ above the ground on reaching the ground it pierces through a distance ‘s’ and then stops finally. If R is the average resistance offered by ground then
\({{W}_{g}}+{{W}_{R}}={{K}_{f}}-{{K}_{i}}\)
\(mg\left( h+s \right)+\left( -R.S \right)=0\)
\(R=mg\left( 1+\frac{h}{s} \right)\)
Here time of penetration is given by impulse equation
\(\left( R-mg \right)t=0-m\sqrt{2gh}\)
\(t=\frac{2s}{\sqrt{2gh}}\)
Ø A body of mass ‘m’ is initially at rest. By the application of a constant force, its velocity changes to V0 in time to the kinetic energy of the body at time ‘t’ is
\(W=\Delta K\)
\(W={{K}_{f}}-{{K}_{i}}\)
\(W={{K}_{f}}-0\)
\({{K}_{f}}=W=mas\)
\(\left( \because w=mas \right)\)
\(=ma\left( \frac{1}{2}a{{t}^{2}} \right)\)
\({{K}_{f}}=\frac{1}{2}m{{a}^{2}}{{t}^{2}}\)
Since \(a=\frac{{{v}_{o}}}{{{t}_{o}}}\)
\(\because {{K}_{f}}=\frac{1}{2}mv_{0}^{2}{{\left( \frac{t}{{{t}_{0}}} \right)}^{2}}\)
Ans: 3/5 m
2. A body of mass 0.5kg travels in a stright line with velocity where what is the workdone by the net force during it’s displacement from x=0 to x=2m.
Ans: 50J
A body is said to be in tanslatory equilibrium, if net force acting on the body is zero
i.e., \(\overline{{{F}_{net}}}=0\)
If the forces are conservative \(F=-\frac{dU}{dr}\) and for equilibrium \(F=0\) , so \(-\frac{dU}{dr}=0\,\,or\,\,\frac{dU}{dr}=0\)
at equilibrium position slope of U-r graph is zero or the potential energy is optimum (maximum or minimum or constant)
There are three equilibriums they are
(i) Stable equilibrium
(ii) Unstable equilibrium
(iii) Neutral equilibri
Stable Equilibrium | Unstable equilibrium |
Neutral Equilibrium |
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Illutrations:
1) The mass of a simple pendulum bob is 100 gm. The length of the pendulum is 1 m. The bob is drawn aside from the equilibrium position so that the string makes an angle of 60° with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be
Ans: 0.49 J
The time of doing work is called power. Power or average power is given by
\({{P}_{avg}}=\frac{work\ done}{time}\)
Power is a scalar
SI Unit : watt(W) (or) J/s
C.G.S Unit : erg/sec
Other Units : kilo watt, mega watt and horse power
One horse power (H.P)=746 watt
\(P=\underset{\Delta t\to .0}{\mathop{Lt}}\,\left( \frac{\Delta W}{\Delta t} \right)\)
It is also calculated by \({{P}_{avg}}=FV\cos \theta =\overrightarrow{F}.\overrightarrow{V}\)
Relation Between \({{P}_{avg}}\) and \({{P}_{ins}}\)
\({{P}_{avg}}=\frac{W}{t}=\frac{m{{v}^{2}}}{2t}\)
\(=\frac{1}{2}mv\left( \frac{v}{t} \right)\)
\(=\frac{1}{2}mav\)
\(=\frac{1}{2}\overrightarrow{F}.\overrightarrow{V}\)
\({{P}_{avg}}=\frac{1}{2}{{P}_{inst}}\)
The area under graph gives work done \(P=\frac{dw}{dt}\)
\(\therefore W=\int{P.dt}\)
The slope of w-t curve gives instantaneous power \(P=\frac{dw}{dt}=\tan \theta\)
Ø the power of a machine gun firing ‘n’ bullets each of mass ‘m’ with a velocity ‘v’ in a time intervel
t’ is given by \(P=\frac{nm{{v}^{2}}}{2t}\)
Ø A crane lifts a body of mass M with a constant velocity V from the ground, its power is \(P=Fv=mgv\)
Ø Power of lungs of a boy blowing whistle is \(P=\frac{1}{2}\) ( mass of air blown per sec) v2
Ø Power of a heart pumping blood is P = pressure X volume of blood pumped per sec.
Ø A conveyor belt is moving with a constant speed ‘v’ horizontally and gravel is fed on it at a rate of \(\frac{dm}{dt}\) Then \(F=V\frac{dm}{dt}\).Then additional power required to drive the belt is, \(P=FV={{V}^{2}}\frac{dm}{dt}\)
Ø When a liquid of density `\(\rho \)` ’ coming out of a hose pipe of area of cross section ‘A’ with a velocity ‘v’ and strikes the wall normally and stops dead. Then power exerted by the liquid is \(P=\frac{1}{2}\frac{m{{v}^{2}}}{t}=\frac{1}{2}\rho A{{v}^{3}}\)
Ø A vehicle of mass ‘m’ is driven with constant acceleration along a straight level road against a constant external resistance ‘R’ when the velocity is ‘v’ power of engine is \(P=Fv\)
\(P=\left( R+ma \right)v\)
Ø If p is a rated power of a device and x% is its efficiency, useful power is (output power) \({{P}^{1}}=\frac{x}{100}.P\)
Ø If a motor lifts water from a well of depth ‘h’ and with a velocity ‘v’ then power of the pump \(P=\frac{mgh+\frac{1}{2}m{{v}^{2}}}{t}\)
Ø If a body of mass ‘m’ starts from rest and accelerated uniformly to a velocity in a time , then the work done on the body in a time ‘t’ is given by
\(W=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{{{V}_{o}}t}{{{t}_{o}}} \right)}^{2}}\)
\(\therefore a=\frac{{{V}_{0}}}{{{t}_{o}}},\,\,\,\,\,\,\ \nu =at=\left( \frac{{{V}_{0}}}{{{t}_{o}}} \right)t\)
Instantaneous power, \(P=FV=mav\)
\(\therefore P=m\frac{{{V}_{0}}}{{{t}_{o}}}\left( \frac{{{V}_{0}}}{{{t}_{o}}} \right)t\)
\(\therefore P=m\frac{v_{0}^{2}}{t_{0}^{2}}.t\)
Ø A motor pump is used to deliver water at a certain rate from a given pipe. To obtain ‘n’ times water from the same pipe in the same time by what amount (a) force and (b) power of the motor should be increased.
If a liquid of denisty `\(\rho \)` is flowing through a pipe of cross section ‘A’ at speed ‘v’ the mas coming out per second will be \(\frac{dm}{dt}=AV\rho \)
To get ‘n’ times water in the same time
\({{\left( \frac{dm}{dt} \right)}^{1}}=n\left( \frac{dm}{dt} \right)\)
\(\therefore A'V'\rho '=n\left( AV\rho \right)\)
As the pipe and liquid are same \({{\rho }^{|}}=\rho \)
\(A'=A,V'=nv\)
Ø Now as \(\frac{F'}{F}=\frac{{{V}^{|}}{{\left( \frac{dm}{dt} \right)}^{1}}}{V\left( \frac{dm}{dt} \right)}=\frac{\left( nv \right)\left( n.\frac{dm}{dt} \right)}{v\left( \frac{dm}{dt} \right)}={{n}^{2}}\)
Ø and as \(P=FV\)
\(\frac{P'}{P}=\frac{F'V'}{FV}=\frac{\left( {{n}^{2}}F \right)\left( nv \right)}{FV}={{n}^{3}}\)
to get ‘n’ times of water force must be increased \({{n}^{2}}\) times while power times.
times while power times.
Conceptual understanding
1)A body starts form rest and acquires a velocity V in time T. The instantaneous power delivered to the body in time ‘t’ is proportional to
Ans: \(\frac{{{V}^{2}}}{{{T}^{2}}}t\)
Illutrations:
Ans 500
Ans: 9 kW
suppose only conservative forces operate on a system of particles and potential energy U is difined correspoding to these forces. There are either no other forces or the workdone by them is zero
\({{U}_{f}}-{{U}_{i}}=-W\)
from work energy theorem \(W={{k}_{f}}-{{k}_{i}}\)
\(\therefore {{U}_{f}}-{{U}_{i}}=-\left( {{k}_{f}}-{{k}_{i}} \right)\)
\(\therefore {{U}_{f}}+{{k}_{f}}={{U}_{i}}+{{k}_{i}}\)
The sum of the potential energy and the kinetic energy is called the total mechnanical energy.
ØTotal mechanical energy of a systme remains constant, if only conservative forces are acting an a system of pasticles and the workdone by all other forces in zero. This is clled the conservation of mechanical energy.
Ø A body is projected vertically up from the ground. When it is at height ‘h’ above the ground, its potential and kinetic energy are in the ratio x : y. If H is the maximum height reached by the body, then
\(\frac{x}{y}=\frac{h}{H-h}\) or \(\frac{h}{H}=\frac{x}{x+y}\)
Vertical Circular Motion with Variable Speed:
Consider a body of mass ‘m’ tide at one end of a string of length ‘r’ and is whirled in a vertical circle by fixing the other end at ‘O’. Let V1 the velocity of the body at the lowest point.
Ø Velocity of the body at any point on the verticle circle
\({{V}_{\theta }}^{2}={{V}_{1}}^{2}-2gh\) but h =\(r\left( 1-\cos \theta \right)\)
\({{V}_{\theta }}^{2}={{V}_{1}}^{2}-2gr\left( 1-\cos \theta \right)\)
.\({{V}_{\theta }}=\sqrt{{{V}_{1}}^{2}-2gr\left( 1-\cos \theta \right)}\)
If V2 is the velocity of the body at highest point \(\theta ={{180}^{0}}\)
\({{V}_{2}}=\sqrt{{{V}_{1}}^{2}-2gr\left( 1+1 \right)}\)
\({{V}_{2}}^{2}={{V}_{1}}^{2}-4gr\)
Tension in the string at any point :
Let \({{T}_{\theta }}\) be the tension in the string when the string makes an angle \(\theta \) with verticle
\({{T}_{\theta }}=\frac{m{{V}_{\theta }}^{2}}{r}+Mg\cos \theta \)
At the lowest point \(\theta ={{0}^{0}}\) the tension in the string is
TL=\(\frac{m{{V}_{1}}^{2}}{r}+Mg\) (maxmum)
At the highest point \(\theta ={{180}^{0}}\)
The tension in the string is
TH = \(\frac{m{{V}_{2}}^{2}}{r}-Mg\) (minimum)
\({{T}_{\left( hor \right)}}=\frac{mV_{horz}^{2}}{r}\)
The difference in maximum and minimum tensionin the string is
Tmax–Tmin = \(\frac{mV_{1}^{2}}{r}+mg-\frac{mV_{2}^{2}}{r}+mg\)
\(=\frac{m}{r}\left( V_{1}^{2}-V_{2}^{2} \right)+2mg\)
\(=\frac{m}{r}\left( 4gr \right)+2mg\)= 4mg + 2 mg = 6 mg
ØRatio of maximum tension to minimum tension in the string is
\(\\frac{{{T}_{\max }}}{{{T}_{\min }}}=\frac{\frac{mV_{1}^{2}}{r}+mg}{\frac{mV_{2}^{2}}{r}-mg}\)
\(=\frac{V_{1}^{2}+rg}{V_{2}^{2}-rg}\left( V_{2}^{2}=V_{1}^{2}-4gr \right)\)
Ø When the particle is at ‘P’ as shown in the fig.
a) Tangential force acting on the particle is \({{F}_{t}}=-mg\sin \theta \)
Tangential acceleration \({{a}_{t}}=-g\sin \theta \)
b) Centribetal force acting on the particle is \({{F}_{c}}=\left( \frac{m{{V}_{\theta }}^{2}}{r} \right)={{T}_{\theta }}-mg\cos \theta \)
Centripetal acceleration \({{a}_{c}}=\frac{V_{\theta }^{2}}{r}\)
c) Net acceleration of the particle at the point ‘P’ is \(a=\sqrt{a_{t}^{2}+a_{c}^{2}}\)
Angle made by net force or net acceleration with centripeta component \(\phi \).is and \(\tan \phi =\frac{{{F}_{t}}}{{{F}_{c}}}\)
d) The net force acting on the particle at point ‘P’
F = \(\sqrt{F_{t}^{2}+F_{c}^{2}}\)
Condition for looping a loop in vertical circular motion:
We know that \({{T}_{2}}=\frac{mv_{2}^{2}}{r}-mg\)
The body will complete the vertical circular path when tension at heighest point is such that
\({{T}_{2}}\ge 0\) \(\frac{mv_{2}^{2}}{r}-mg\ge 0\)
\({{V}_{2}}_{\min }=\sqrt{gr}\)
Hence the minimum speed at highest point to just complete the vertical circle is \(\sqrt{gr}\)
From law of conservation of mechanical energy
Total energy at lowest point A = total energy at highest point B
\({{U}_{A}}+K{{E}_{A}}={{U}_{B}}+K{{E}_{B}}\)
\(\frac{1}{2}mv_{1}^{2}=2mgr+\frac{1}{2}mgr\left[ \because {{V}_{2}}=\sqrt{gr} \right]\)
\(=\frac{5}{2}mgr\)
\({{V}_{1}}=\sqrt{5gr}\)
For the body to continue along circular path the critical velocity at lowest point is \(\sqrt{5gr}\)
Conceptual understanding
1) For a body thrown vertically upwards, its direction of motion changes at the point where its total mechanical energy is
Ans: equal to the potential energy.
Illutrations:
1) The length of a ballistic pendulum is 1 m and mass of its block is 0.98 kg. A bullet of mass 20 gram strikes the block along horizontal direction and gets embeded in the block. If block + bullet completes vertical circle of radius 1m, the striking velocity of bullet is
Ans: 350 m/s
2)A body of mass m is rotated at uniform speed along vertical circle with help of light string. If are tensions in the string when the body is crossing highest and lowest point of vertical circle respectively then following expression is correct.
Ans: \({{T}_{2}}-{{T}_{1}}=6mg\)
Potential energy of a body is the enrgy possessed by a body by virtue of its position or configuration in the gravitational field.
Penergy is defined only for conservative forces. It does not exist for non-conservative forces. In case of conservative forces.
\(F=-\left( \frac{dU}{dr} \right)\) U is P.E
\(\therefore dU=-\bar{F}.\overline{dr}\)
\(\int\limits_{{{u}_{1}}}^{{{u}_{2}}}{dU}=-\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\overline{dr}}\)
\({{U}_{2}}-{{U}_{1}}=-\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\,d\overline{r}}=-w\)
\(\left( \because \int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\overline{F}.\,d\overline{r}=w} \right)\)
\(\therefore \,\,\,If\,\,{{r}_{1}}=\infty \,,\,{{U}_{1}}=0\)
\(\therefore \,\,U=-\int\limits_{\infty }^{r}{\overline{F}.d\overline{r}=-w}\)
P.E can be +ve or -ve or can be zero.
P.E depends on frame of reference.
Examples: water stored in a dam possesses P.E A streteched bow possesses P.E
Gravitational PE:
A body of mass m is at a height h above the ground. Its PE \(U=\frac{mgh}{1+\frac{h}{R}}\)
where R is radius of the earth, g-acceteration due to gravity.
If h <<< R then , \(\frac{h}{R}<<<1\)we can neglect \(\frac{h}{R}\)
U= mgh
A moving body may or may not have P.E
If P.E of the system is negative then the particles are bounded to the system.
If PE of the system is positive the particles are not bounded to the system.
Ø The absoulte value of PE has no physical significance.
Spring Forces:
Spring force is an example of a variable force which is conservative. The figure shows one end of a massless spring attached to a rigid vertical support and the other end to a block of mass ‘m’ which can move on a smooth horizontal surface.
t Lex = 0 denote the position of block when the spring is at its natural length. In an ideal spring, the spring force \({{F}_{s}}\) is directly proportional to ‘x’
where x is the displacement of the block from equilibrium position. This can be expressed as
\({{F}_{s}}=-Kx\)
The constant K is clled the spring constant.
To calculate the workdone on the block by the spring force as the block moves from undeformed position
\(x=0\,to\,x={{x}_{1}}\)
\(dw=\overline{F}.d\overline{x}=-kxdx\)
\(\int{dw=\int\limits_{0}^{{{x}_{1}}}{-kxdx}}\)
\(\therefore \,\,{{W}_{1}}=-\frac{1}{2}k\left( {{x}^{2}} \right)_{0}^{{{x}_{1}}}=-\frac{1}{2}kx_{1}^{2}\)
The workdone by the spring force is always negative. similar is the result of the workdone by spring force
when it is compressed through \({{x}_{1}}\) from its natural length \(w=-\frac{1}{2}kx_{1}^{2}\)
If the block moves from \(x={{x}_{1}}\,to\,x={{x}_{2}}\)
workdone by spring force is \(w=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{-kxdx}\)
\(w=\,\frac{1}{2}k\left( x_{1}^{2}-x_{2}^{2} \right)=\frac{1}{2}kx_{1}^{2}-\frac{1}{2}kx_{2}^{2}\)
Conceptual understanding
Ans: electrostatic
Ans: the change in potential energy of the system.
Illutrations:
Ans: 200 N
2.The elastic potential energy of a stretched spring is given by E = 50x2 where x is the displacement in meter and E is in joule, then the force constant of the spring is
Ans: 100 N/m².