Showing posts with label Numerical Problems and Solutions. Show all posts
Showing posts with label Numerical Problems and Solutions. Show all posts

Numerical Problems and Solutions of Power

Formulas and Solutions to Numerical Problems of Power

Important Formula of Power:

1.) $P=\frac{Work(W)}{Time(t)}$

2.) $P=\frac{F.s}{t}$

3.) $P=\frac{F \: s \: cos \theta}{t}$

4.) $P=\frac{F \: s }{t} \qquad \left( \because \theta =0^{\circ} \right)$

5.) $P=F \: v \qquad \left( \because v = \frac{s}{t} \right)$

Numerical Problems and Solutions


Q.1 A woman pulls a bucket of water of mass $5 Kg$ from a well which is $10 m$ deep in 10 sec. Calculate the power used by her $(g=10 \: m/sec^{2})$.

Solution:
Given that:
The mass of bucket of water $(m)=5 \: Kg$ Depth of well $(h)=10 \: m$

The time taken to pull a bucket from the well $t= 10 \: sec $

The value of gravitational acceleration $g=10 \: m/sec^{2}$

The power used $(P)=?$


Now the power used by her:


$P=\frac{W}{T}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the equation of power:

$P=\frac{5 \times 10 \times 10}{10}$

$P=50 \: Joule/sec=50 W$

Q.2 A man whose mass is $50 \: Kg$ climbs up $30$ steps of the stairs in $30 \: Sec$. If each step is $20 \: cm$ high, Calculate the power used in climbing the stairs $(g=10 \: m/sec^{2})$

Solution:
Given that:
Mass of a man $(m) = 50 \: Kg$

Climbs up the number of Steps $(N) = 30$

The time is taken to climb up the 30 steps $(t)= 30 \: sec$

The length or height of one step $(l)=20 cm \: = \: 0.20 \: m$

The total length or height of 30 steps $h=30 \times 0.20 \: m$

The power used in climbing the stairs=?

Now, from the equation of power:

$P=\frac{W}{t}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the above equation:

$P=\frac{50 \times 10 \times 30 \times 0.20 }{30}$

$P=100 \: Joule/sec = 100 \: W$

Q.3 A horse exerts a pull of $300 N$ on a cart so that the horse-cart system moves with a uniform speed of $18 Km/h$ on a level road. Calculate the power in watts developed by the horse and also find its equivalent in horsepower.

Solution:
Given that:
The horse exerts the pull, i.e, force $(F)=300 N$

The unifrm speed of the horse-cart $(v)=18 Km/h$

So the distance moved in $(s)=18 Km=18000m$

The time $(t)=1h= 60 \times 60 sec$

The power developed by the horse$(P)=?$

From the equation of the power, the power developed by a horse in one hour:

$P=\frac{W}{t}$ $P=\frac{F.s}{t}$

Now substitute the given values in the above equation:

$P=\frac{3000 \times 18000}{60 \times 60}$

$P=1500 W$

$P=\frac{1500}{746}$

$P=2 \: hp$

Q.4 A man weighing $60Kg$ climbs up a staircase carrying a load of $20 Kg$ on his head. The staircase has 20 steps, each of height $0.2m$. If he takes $10 sec$ to climb, find his power.

Solution:
Given that:
The weight of man $(m_{1})=60Kg$

The weight of load $(m_{2})=20Kg$

The number of steps in the staircase $(N)=20$

The height of each step $(H)=10s$

The total mass of the man and load $(m)=m_{1}+m_{2}= 60+20=80 Kg$

The total heght of stair case $(h)=N \times H = 20 \times .2= 4 m$



The power of man: $P=\frac{W}{t}$

$P=\frac{mgh}{t}$

$P=\frac{80 \times 9.8 \times 4}{10}$

$P=313.6 W$

Q.5 A car of mass $2000 Kg$ and it is lifted up a distance of $30m$ by a crane in $1 \: min$. A second crane does the same job as the first crane in $2 \: min$. Do both cranes consume the same or different amounts of fuel? Find the power supplied by each crane? Neglecting power dissipation against friction.

Solution:
Given that:
The mass of car $(m)=2000 Kg $

The lifted up distance $(h)=30m$

The time taken by first crane $(t_{1}= 1\: min)$

The time taken by second crane $(t_{2}= 2\: min)$

The work done by each crane:

$W=mgh$

$W=2000 \times 9.8 \times 30$

$W=5.88 \times 10^{5} J$

As both the cranes do the same amount of work, both consume the same amount of fuel.

The power supplied by the first crane:

$P_{1}=\frac{mgh}{t_{1}}$

$P_{1}=\frac{2000 \times 9.8 \times 30}{60}$

$P_{1}=9800 W$

The power supplied by the second crane:

$P_{2}=\frac{2000 \times 9.8 \times 30}{1.2}$

$P_{2}=4900 W$

Q.6 The human heart discharges $75 \: mL$ of blood at every beat against a pressure of $0.1 m$ of Hg. Calculate the power of the heart assuming that the pulse frequency is $80$ beats per minute. Density of $Hg=13.6 \times 10^{3} Kg/m^{3}$.

Solution:
Given that:
The volume of blood discharge per beat $(V)=75 \: mL = 75 \times 10^{-6} m^{-3} $

The pressure of blood $(P)=0.1 m \: of \: Hg$ i.e.

$P= 0.1 \times 13.6 \times 10^{3} N-m^{-2} \quad (\because P=\rho g h)$

The work done per beat $=PV$

The workdone in 80 beats$W =80 \times PV$

The power of heart:

$P=\frac{80 \times PV}{60}$

$P=\frac{80 \times 0.1 \times 13.6 \times 10^{3} \times 75 \times 10^{-6} \times 9.8}{60}$

$P=1.33 W$

Q.7 A machine gun fires $60$ bullets per minute with a velocity of $700 m/sec$. If the mass of each bullet is $50 g$, then find the power developed by the gun.

Solution:
Given that:
The mass of one bullet is $M=50 g$

The number of bullets $N=60$

The mass of $60$ bullets $m=M \times N= 50\times 60 = 300g= 3Kg$

The velocity of the bullet $v=700 m/sec$

The time take to fire $60$ bullets $t=1 \: min = 60 sec$

The power developed by gun:

$P=\frac{W}{t}$

According to the work-energy theorem:

$W=\Delta K$

$P=\frac{\Delta K}{t} $

$P=\frac{m v^{2}}{2t} \quad( \because K=\frac{mv^{2}}{2})$

$P=\frac{3 \times (700)^{2}}{2 \times 60} $

$P=12250 W $

Numerical Problems and Solutions of Centripetal Force and Centripetal Acceleration

Solutions to Numerical Problems of the Centripetal Force and Centripetal Acceleration

Numerical Problems and Solutions

Q.1 An object of mass $0.20 \: Kg$ is being revolved in a circular path of diameter $2.0 \: m$ on a frictionless horizontal plane by means of a string. It performs $10$ revolutions in $3.14 \: s$. Find the centripetal force acting on an object.

Solution:
Given that:
Mass of body $(m)=0.20 \: Kg$
Diameter of Circular path $(d)=2.0: m$
Number of revolutions $(n)=10$
Time taken to complete $(10)$ revolution$(t)=3.14 s$
The centripetal force acting on a body $(F)=?$
Now the centripetal force:
$F=m r \omega^2$
$F=m r \left( \frac{2 \pi n}{t} \right)^2$
Now Substitute the given values in the centripetal force formula:
$F=0.20 \times 1 \left( \frac{2 \times 3.14 \times 10}{3.14} \right)^2$
$F=0.8 N$

Q.2 A car of mass $1200 Kg$ is moving with a speed of $10.5 ms^{-1}$ on a circular path of radius $20 m$ on a level road. What will be the frictional force between the car and the road so that the car does not slip?

Solution:
Given that:
Mass of car $(m) = 1200 \: Kg$
Speed of car $(v) = 10.5: ms^{-1}$
The radius of circular path $(r)= 20 m$
The frictional force $(F)=?$
The frictional force $(F)$ should be equal to the required centripetal force i.e
$F=\frac{mv^{2}}{r}$
Now Substitute the given values in the centripetal force formula:
$F=\frac{1200 \times (10.5)^{2}}{20}$
$F=6.615 \times 10^{3} N$


Q.3 A string can bear a maximum tension of $50 N$ without breaking. A body of mass $1 kg$ is tied to one end of $2 m$ long piece of the string and rotated in a horizontal plane. Find the maximum linear velocity with which the string would not break.

Solution:
Given that:
Maximum tension on the string without breaking $(F)=50 N$
mass of the a body $(m)=1 Kg$
length of the string $(l)=2 m$
The maximum velocity with which the string would not break $(v)=?$
We know that the centripetal force:
$F=\frac{mv^{2}}{r}$
$v=\sqrt{\frac{F \: r}{m}}$
Now Substitute the given values in the centripetal force formula:
$v=\sqrt{\frac{50 \times 2 2}{1}}$
$v=10 ms^{-1}$


Q.4 The moon revolves around the earth in $2.36 \times 10^{6} s$ in a circular orbit of radius $3.85 \times 10^{5} Km$. Calculate the centripetal acceleration produced in the motion of the moon.

Solution:
Given that:
The radius of circular orbit of the moon $(r)=3.85 \times 10^{5} Km = 3.85 \times 10^{8} m$
Time taken to complete one revolution $(T)=2.36 \times 10^{6} s$
Centripetal acceleration produced in the motion of the moon $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{2.36 \times 10^{6}} \right) \times 3.85 \times 10^{8} $
$a=2.73 \times 10^{-3} ms^{-2} N$


Q.5 Calculate the centripetal acceleration at a point on the equator of the earth. The radius of the earth is $6.4 \times 10^{6} m$ and it completes one rotation per day about its axis.

Solution:
Given that:
Radius of earth $(r)=6.4 \times 10 ^{6} m$
Time taken to complete one rotation per day about its axis$(T)=24 \times 60 \times 60 s$
centripetal acceleration at a point on equator $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{24 \times 60 \times 60} \right) \times 6.4 \times 10^{6} $
$a=3.37 \times 10^{-2} ms^{-2} N$


Popular Posts

Study-Material













  • Classical world and Quantum world

  • Inadequacy of classical mechanics

  • Drawbacks of Old Quantum Theory

  • Bohr's Quantization Condition

  • Energy distribution spectrum of black body radiation

  • Energy distribution laws of black body radiation

  • The Compton Effect | Experiment Setup | Theory | Theoretical Expression | Limitation | Recoil Electron

  • Davisson and Germer's Experiment and Verification of the de-Broglie Relation

  • Significance of Compton's Effect

  • Assumptions of Planck’s Radiation Law

  • Derivation of Planck's Radiation Law

  • de-Broglie Concept of Matter wave

  • Definition and derivation of the phase velocity and group velocity of wave

  • Relation between group velocity and phase velocity ($V_{g}=V_{p}-\lambda \frac{dV_{p}}{d\lambda }$)

  • Group velocity is equal to particle velocity($V_{g}=v$)

  • Product of phase velocity and group velocity is equal to square of speed of light ($V_{p}.V_{g}=c^{2}$)

  • Heisenberg uncertainty principle

  • Mathematical equation of wave function of a free particle in simple harmonic motion

  • Physical interpretation of the wave function

  • Derivation of time dependent Schrodinger wave equation

  • Derivation of time independent Schrodinger wave equation

  • Eigen Function, Eigen Values and Eigen Vectors

  • Postulate of wave mechanics or Quantum Mechanics

  • Quantum Mechanical Operators

  • Normalized and Orthogonal wave function

  • Particle in one dimensional box (Infinite Potential Well)

  • Minimum Energy Or Zero Point Energy of a Particle in an one dimensional potential box or Infinite Well

  • Normalization of the wave function of a particle in one dimension box or infinite potential well

  • Orthogonality of the wave functions of a particle in one dimension box or infinite potential well

  • Eigen value of the momentum of a particle in one dimension box or infinite potential well

  • Schrodinger's equation for the complex conjugate waves function

  • Probability Current Density for a free particle in Quantum Mechanics

  • Ehrenfest's Theorem and Derivation

  • Momentum wave function for a free particle

  • Wave function of a particle in free state

  • One dimensional Step Potential Barrier for a Particle

























  • Blog Archive