Tag: energy and its forms

Questions Related to energy and its forms

A force applied by the engine of a train of mass 2.05 x $10^{6} kg$ changes its velocity from 5 $ms^{-1}$ to 25  $ms^{-1}$ in 5 minutes. The power of the engine is then

power work and power work, energy and power physics energy and its forms

  1. 1.025 MW

  2. 2.05 MW

  3. 5 MW

  4. 6 MW

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Correct Option: B
Explanation:

Given that,

Mass of train, $m=2.05\times {{10}^{6}}\,Kg$

Time, $t=5\,\min utes=300\,s$

$ v=25\,m/s $

$ u=5\,m/s $

Acceleration,

$ a=\dfrac{v-u}{t} $

$ a=\dfrac{25-5}{300} $

$ a=\dfrac{2}{30}=\dfrac{1}{15}\,m/{{s}^{2}} $

Using equation of motion,

$ {{v}^{2}}-{{u}^{2}}=2as $

$ {{(25)}^{2}}-{{(5)}^{2}}=2\times \dfrac{1}{15}\times s $

$ s=4500\,m $

Power,

$ P=\dfrac{work\,\,done}{time} $

$ P=\dfrac{F\times d}{t} $

$ P=\dfrac{m\times a\times s}{t} $

$ P=\dfrac{2.05\times {{10}^{6}}\times 1\times 4500}{15\times 300} $

$ P=2.05\times {{10}^{6}}\,W $

$ P=2.05\,MW $

A motor lifts $100 kg$ of water in $2 min$ from a well of $60m$ depth then the electric power of the motor is$(Taken g=10 m/s^2)$

power work and power work, energy and power physics energy and its forms

  1. $1000 W$

  2. $750 W$

  3. $1200 W$

  4. $500W$

Show answer
Correct Option: D
Explanation:

A motor lifts $=100kg$ of water

Time $=2min=2\times60=120s$
Depth$=60m$ depth then,
electric power of the motor$=?$
Taking $=10m/s^2$
$P=Power=\cfrac{mgh}{t}\ \quad=\cfrac{100\times10\times60}{120}\ \quad=500W$

A force $'F'$ accelerates a block of mass $'m'$ along a straight line to velocity $'v'$ from rest and displace it through a distance $'s'$. What is the average power developed?

power work and power work, energy and power physics energy and its forms

  1. $\dfrac {v^{2}}{2F}$

  2. $Fv$

  3. $\dfrac {mv^{2}}{2s}$

  4. $\dfrac {Fv}{2}$

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Correct Option: D

The average power required to lift a 100 Kg mass trough a highest of 50 meters in approximatley 50 second would be

power work and power work, energy and power physics energy and its forms

  1. 50 J/s

  2. 5000 J/s

  3. 100 J/s

  4. 980 J/s

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Correct Option: D

A pump ejects $12000kg$ of water at speed of $4m/s$ in $40$ second. Find the average rate at which the pump is working

power work and power work, energy and power physics energy and its forms

  1. $0.24KW$

  2. $2.4KW$

  3. $24KW$

  4. $24W$

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Correct Option: B

A refrigerator transfer 180 joule of energy in one second from temperature $ { -3 }^{ 0 }C$ to $ { 27 }^{ 0 }C.$ Calculate the average power consumed, assuming no energy losses in the process. 

power work and power work, energy and power physics energy and its forms

  1. 18 W

  2. 54 W

  3. 20 W

  4. 120 W

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Correct Option: A

A man pulls a lawn roller with a force of $20\ kgf$. He applies the force at an angle $60^{\circ}$ with the ground through $10\ m$. If he takes $1$ minutes in doing so, calculate the power developed.

power work and power work, energy and power physics energy and its forms

  1. $25\ J$

  2. $16.67\ J$

  3. $34\ J$

  4. $58\ J$

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Correct Option: B

If the average power radiated by the star is $10 ^ { 16 } \mathrm { W }$ , the deuteron supply of the star is exhausted in a time of the order of 

power work and power work, energy and power physics energy and its forms

  1. $10 ^ { 6 }$ seccond

  2. $10 ^ { 14 }$ second

  3. $10 ^ { 12 }$ second

  4. $10 ^ { 16 }$ second

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Correct Option: C

A boat moving with constant speed v in still waters experiences a total frictional force F. The power developed by the boat is

power work and power work, energy and power physics energy and its forms

  1. $\frac{1}{2}Fv$

  2. $Fv$

  3. $\frac{1}{2}Fv^2$

  4. $Fv^2$

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Correct Option: B

A body of mass m is projected at an angle $\displaystyle \theta $ with the horizontal with an initial velocity $\displaystyle v _{0}.$ The average power of gravitational force over the whole time of flight is

power work and power work, energy and power physics energy and its forms

  1. $\displaystyle mg\cos \theta $

  2. $\displaystyle \frac{1}{2}mg\sqrt{u\cos \theta }$

  3. $\displaystyle \frac{1}{2}mgu\sin \theta $

  4. zero

Show answer
Correct Option: D
Explanation:

The projected body again comes back down. Hence the net displacement in the vertical direction will be 0.
$\therefore W=mgh=mg\times 0=0$