Given,
Weight of the car $(W) =1200 \; N$
Coefficient of friction $(\mu) = 0.015$
Velocity $(v) = 72\;km/hr = 20\;m/s$
Horizontal force $(F) = ?$
Power developed by the engine $(P) = ?$
Now we have,
$F = F_k + ma$ [Here, $F_k = \mu_r\;.\;mg$ = frictional force]
If the speed is uniform, then a = 0,
$F = \mu_r\;.\;mg = 0.015 * 1200 = 18\;N$
Therefore, horizontal force needed $(F) = 18\;N$
Also, Power developed by the engine $(P) = F * v$
$= 18 * 20 = 360\;N$
Given,
First lamp = 25 W - 220 V
Second lamp = 100 W - 220 V
Voltage (V) = 220 V
Power Consumed (P) = ?
Now, to calculate the resistance
For first lamp, $P_1 = 25\;W$
Voltage $(V_1) = 220\;V$
$R_1 = $$ \frac{V_1^{2}}{P_1} = \frac{220^2}{25}$$ = 1936\; \Omega$ [∵ $P = \frac{V^2}{R}$]
For second lamp, $P_2 = 100\;W$
Voltage $(V_2) = 220\;V$
$R_2 = $$\frac{V_2^{2}}{P_2} = \frac{220^2}{100}$$ = 484 \; \Omega$
If two lamps are connected in series and joined to 220 V mains, then the current in the circuit (I) is calculated as,
$I = $$\frac{V}{R_1 + R_2} = \frac{220}{1936 + 484}$$ = 0.091 \;A$
Finally,
Power consumed by the first lamp = $I^2*R_1 = (0.091)^2 * 1936 = 16\;W$
Power consumed by the second lamp = $I^2*R_2 = (0.091)^2 * 484 = 4\;W$
The average power absorbed by each light bulb is directly proportional to the amount of light it emits.
Case I:
Let $R_1 = R_2 = R_3 = R_4 = R_5 = 100\; \Omega$
Then total resistance $R = 500\;\Omega$
Now, Power $(P_1) = $$\frac{V^2}{R} = \frac{(220)^2}{500}$$ = 96.8\;\;Watt$
Case II:
Let $R_1 = R_2 = R_3 = R_4 = 100\; \Omega$
Then total resistance $R = 400\;\Omega$
Now, Power $(P_1) = $$\frac{V^2}{R} = \frac{(220)^2}{400}$$ = 121\;\;Watt$
Since $P_2 > P_1$
Therefore the second case with four light bulbs draws more power and will produce more light.
Power ($P$) = $I * E$ .......... (i) (∵ P = I * IR = I$^2$R)The instantaneous value of current and emf are,
$\left.\begin{matrix}
I = I_0 Sin \omega t
\\
E = E_0 Sin (\omega t + \theta)
\end{matrix}\right\}$ .......... (ii)
Power (P) = $I_0$ Sin $\omega t . E_0 $ Sin $(\omega t + \theta)$
= $I_0 . E_0$ Sin $\omega t$ (Sin $\omega t$ Cos $\theta$ + Cos $\omega t$ Sin $\theta)$
= $I_0 . E_0$ (Cos $\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t)$
$dW = P.dt$
or, $dW = I_0 . E_0$ (Cos$\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t)$ .......... (iii)
or, $\int dW = \int_{0}^{T} I_0 . E_0$ (Cos$\theta$ . Sin$^2 \omega t$ + Sin$\theta$ . Cos $\omega t$ . Sin $\omega t) $
$W = \frac{I_0 . E_0}{2} T. Cos\theta$
$P_{av} = \frac{W}{T} = \frac{I_0 . E_0}{2\;*\;T} T. Cos \theta $
$ = \frac{I_0 . E_0}{2} Cos\theta$
$ = \frac{I_0}{\sqrt{2}} * \frac{E_0}{\sqrt{2}} * Cos\theta$
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} * Cos \theta$ .......... (iv)
Case I: Power Consumption across R, If $\theta = 0$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} $
Case II: Power Consumption across L, If $\theta = \frac{\pi}{2}$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} * Cos(\frac{\pi}{2}) = 0 $
Case III: Power Consumption across C, If $\theta = \frac{-\pi}{2}$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} Cos(\frac{-\pi}{2}) = 0 $
Case IV: Power Consumption across RL, If $\theta = \frac{R}{Z}$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{X_{L}{^2} + R^2}})$
Case V: Power Consumption across RC, If $\theta = \frac{R}{Z}$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{X_{C}{^2} + R^2}})$
Case VI: Power Consumption across LCR, If $\theta = \frac{R}{Z}$; Then,
$\Rightarrow$ $P_{av} = I_{rms} * E_{rms} (\frac{R}{\sqrt{{X_{L} - X_{C}){^2} + R^2}}}) $
Net work on an object causes a change in the kinetic energy. The work done in a body due to a force is equal to the change in its kinetic energy.
i.e. Work done = Change in Kinetic Energy
Consider a body of mass $M$, moving with a initial velocity $u$ when a constant force $F$ is acts on it. Let final velocity of the body becomes $v$ after covering a distance $d$ in a direction of force. Then, work done by the force is given by
$W = F.d$ .......... (i)
From the Newton's law of Motion:
$F = ma$ .......... (ii)
From equation (i) and (ii)
$F = ma.d$ .......... (iii)
And from the Kinematics, we know the formula:
$v^2 = u^2 + 2ad$
$d = \frac{v^2 - u^2}{2a}$ .......... (iv)
From the equation (i), (iii) and (iv), we get
$W = ma\;(\frac{v^2 - u^2}{2a})$
$= m\;(\frac{v^2}{2} - \frac{u^2}{2})$
$= \frac{1}{2}\,mv^2 - \frac{1}{2}\,mu^2 $
= Final K.E - Initial K.E.
∴ Work done = Change in Kinetic Energy
So, the work done on moving the body by the force is equal to the increase in its kinetic energy.
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If work is done faster, power is higher. If work is done slower, power is smaller.
Power is the rate at which work is done. It is also called the energy is transferred in a unit of time. If work is done faster, power is higher. If work is done slower, power is smaller. For example:
In the race between the tortoise and the rabbit, the rabbit has more power and accelerated faster. But the tortoise did the same work and covered the same distance in a much longer time. The tortoise showed less power.
Mathematically,
Power is defined as the rate at which the work is done.
$Power\;(P) = \frac{Work\; done}{time\; taken} = \frac{W}{t}$ .......... (i)
or,
$Power\;(P) = \frac{W}{t} = \frac{\vec{F}.\;\vec{s}}{t} = \vec{F}.\vec{v}$ .......... (ii)
Where, $F$ = Force and $v$ = velocity.
Thus, power of an agent measures how fast it can do the work. It is a scalar quantity (having no direction). In SI system, the unit of power is Joule / Second ($J/S$) or Watt ($W$).
Another unit of power is horse-power ($h.p$). 1 h.p. = 746 Watt.
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Potential Energy:
"Energy that is stored and waiting to be used later."
Potential energy is due to the position of an object. It is not apparent until released. The stored energy of position is referred to as potential energy. Examples:
There are three types of potential energy:
1) Gravitational Potential Energy:
On earth, there is always have the force of gravity acting on us. When we separate from the earth's surface, we have the potential (stored) energy. This stored energy is called gravitational potential energy.
Consider a body of mass $m$ held stationary at height $h$ from the ground. The energy is released when the object falls to the ground, is equal to the work done by the body falling through this height. i.e.
P.E. = Work done by the force (or, Work done) in distance $h$.
or, P.E. = F * h
∴ P.E. = $m\,g\,h$ .......... (i)
Where, $m$ = mass, $h$ = height, $g$ = gravitational field strength ($9.8\; N/kg$ on Earth).
The amount of gravitational potential energy an object on earth has depends on its Mass ($M$) and height above the ground ($h$).
2) Elastic potential energy:
Potential energy due to compression or expansion of an elastic objects. It can be found in rubber band and springs.
For certain springs (those obeys the HOOK'S Law), the amount of force is directly proportional to the amount of stretch or compression ($x$) is,
F$_{spring} = k . x$ .......... (i)
Where, $k$ is proportionality constant called spring constant.
In terms of potential energy, the special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or compression) and the spring constant (k). The equation is,
P.E = $\frac{1}{2}k\,x^2$ .......... (ii)
Where, $k$ = spring constant
$x$ = amount of compression or stretch (relative to equilibrium position).
3) Chemical potential energy:
Potential energy stored with in a chemical bonds of an object. It is released when chemical reactions take place.
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Energy is a central concept in Physics. It can be defined as the capacity for doing work. But, it is important to understand that just because energy exists, not necessarily available to do work.
Energy can not be create or destroyed, but it can change from one form to another form.
There are many types of energy such as: Mechanical energy, chemical energy, light energy, thermal energy, electrical energy etc. Mechanical energy further divided into two parts called: Kinetic Energy and Potential Energy. Here, we mainly focused on these two energy.
Kinetic Energy:
Kinetic energy is the energy of motion. It is a capacity of doing work due to its motion. There are many forms of kinetic energy - vibrational (energy due to vibrational motion), rotational (energy due to motion in circle) and translation (energy due to motion in straight line). Here, we will focus upon translation kinetic energy to keep matters simple.
In classical mechanics, the kinetic energy of an (translation) object has depends upon mass (m) and velocity (v) of an object. i.e.
$K.E. = \frac{1}{2}mv^2$
In relativistic mechanics, this is a good approximation only when $v$ is much less than the speed of light.
The amount of kinetic energy an object has depends on the mass and speed of the object.
All moving object have the kinetic energy. The faster object has more speed and has more kinetic energy.
» Calculation of Kinetic Energy:
Consider a body of mass $m$ lying on a horizontal surface. After applying a force $F$, the body travel a distance $d$ with a moving velocity $v$.
As we know from the definition of work:
W = force * distance = $F * d$ .......... (i)
From the Newton's law of motion:
$F = ma$ .......... (ii)
And from the Kinematics, we know the formula:
$v^2 = u^2 + 2ad$ .......... (iii)
If the initial velocity of a body is at rest, so the initial velocity is zero ($u = 0$), then equation (iii) becomes,
$ad = \frac{v^2}{2}$ .......... (iv)
From equation (i), (ii) and (iii), we have
$W = F * d = ma *d $
$= m * ad = m * \frac{v^2}{2}$
Thus,
∴ $W = K.E. = \frac{1}{2}mv^2$ .......... (v)
Where, m - mass and v = velocity of a body.
This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means, for a two fold increase in speed, the kinetic energy will increase by the factor four and so on.
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In general, the work done by a force is equal to the product of the force and the displacement of its point of application in the direction of the force.
Work (symbol: W): is defined as transfer of energy by force from one object to another object. If one object transfers energy to a another object , then said that the first object does work on the second object.
Suppose, a force is applied on an object and object is displaced from its position due to the application of the force. It is said to be work done by the force. There are three major component to work: 'Force, Displacement and Cause'. In order to force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement (the physics classroom).
Mathematically,
Work is defined as the product of force and displacement of an object in the direction of force.
i.e. $W = \vec{F}.\vec{d}$ .......... (i)
Where; W = Work done;
F = Force;
d = Displacement.
If $\theta$ is the angle between $\vec{F}$ and $\vec{d}$ then,
Work done by the force (W) = Force * distance moved in direction of force.
$W = F . d\; Cos \theta$ .......... (ii)
If the force and and displacement are in the same direction, then $\theta = 0^{0} \;\;(i.e.\;\; Cos0^{0} = 1)$ From the equation (ii), we have
∴ $W = Fd$ .......... (iii)
Note: Here, no work is done by $F\;Sin\theta$ component because displacement in the direction of this component is zero.
» Unit of work is joule ($J$), in SI system. It's dimensional is [$M\;L^2\; T^{-2}$].
Relation between Joule and erg is,
$1\;J = 1\;N * 1\;m = 10^5\; dyne * 100\;cm = 10^7\;erg $
Work is a scalar quantity, it has no property of direction but only magnitude. When the force is one Newton and the distance moved is one meter, then the work done is one Joule.
Another Example: A force of 50 N moving through a distance of 10 m does 50 * 10 = 500 Joule of work. (This is also a measure of the energy transferred to the object).
» The angle measure is defined as the angle between the force and the displacement.
Case (i): The force vector and the displacement vector are in the same direction. That is, the angle between $F$ and $d$ is $0$ degrees. A force acts rightward upon an object as it is displaced rightward.
Case (ii): The force vector and the displacement vector are in opposite direction. That is, the angle between $F$ and $d$ is $180$ degrees. A force acts leftward upon an object that is displaced rightward.
Case (iii): The force vector and the displacement vector are at right angles to each other i.e. angle between $F$ and $d$ is $90$ degrees. If the force is perpendicular to the direction of displacement , work done is zero.
Example: A waiter who carried a tray full of meals above his hand. The force supplied by the waiter on the tray is an upward and the displacement of the tray is a horizontal with a waiter speed. In this case, the angle between the force and displacement is $90$ degrees. If we calculated the work done by waiter on the tray, the result would be zero.
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