Tag: work and power

Questions Related to work and power

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

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Correct Option: D
Explanation:

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

A time varying power $P=2t$ is applied on a particle of mass $m$. Find average power over a time interval from t=0 to t=t :

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

  1. $\displaystyle P _{av}= t$

  2. $\displaystyle P _{av}= 2t$

  3. $\displaystyle P _{av}= 4t$

  4. $\displaystyle P _{av}= 8t$

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Correct Option: A
Explanation:
Average power over a time interval $t=0$ to $t=t$ is 
$\dfrac{\int _0^tPdt}{\int _0^tdt}$
$=\dfrac{\int _0^t 2tdt}{t}$
$=\dfrac{t^2}{t}=t$

A car of mass 1000 kg accelerates from rest to $100 : km : h^{-1}$ in 5 seconds. What is the average power of the car ?

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

  1. $7.71\times 10^5 W$

  2. $7.71\times 10^4 W$

  3. $15.42\times 10^4 W$

  4. $15.42\times 10^5 W$

Show answer
Correct Option: B
Explanation:


Mass of the car m$=1000\ kg$
Initial velocity of car ${v} _{1}=0$
Final velocity of car ${v} _{2}$$=100km/h=100\times\dfrac{5}{18}m/s=27.78m/s$
Average power of car$=$change in kinetic energy per unit time$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}-\dfrac{1}{2}m{{v} _{1}}^{2}}{t}$,since ${v} _{1}=0$
So,average power of car$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}}{t}$
                                 $=\dfrac{\dfrac{1}{2}\times1000\times{27.78}^{2}}{5}$
                                 $=7.71\times{10}^{4}W$
                                

A particle of mass $m$ is lying on smooth horizontal table. A constant force $F$ tangential to the surface is applied on it. Find average power over a time interval from $t=0$ to $t=t$.

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

  1. $\displaystyle \frac{F^{2}t}{3m}$

  2. $\displaystyle \frac{F^{2}t}{2m}$

  3. $\displaystyle \frac{2F^{2}t}{3m}$

  4. $\displaystyle \frac{3F^{2}t}{2m}$

Show answer
Correct Option: B
Explanation:

Average power of a time interval $t=0$ to $t=t$


is $\dfrac{\int^t _0 Fvdt}{ _0^t\int dt}$


$=\dfrac{F\int _0^t vdt}{t}$

$=\dfrac{F\int _0^t \dfrac{F}{m}tdt}{t}$

$=\dfrac{F^2t}{2m}$

A block of mass $1$ $kg$ starts moving with constant acceleration $\displaystyle a= 4m/s^{2}.$. Find the average power of the net force in a time interval from $t=0$ to $t=2s$.

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

  1. $16 W$

  2. $1.6 W$

  3. $15 W$

  4. $1.5 W$

Show answer
Correct Option: A
Explanation:
Velocity of an object after time $t$ is given by $v=at$
Average power from time $t _1$ to $t _2$

$=\dfrac{\int _{t _1}^{t _2}Fvdt}{\int _{t _1}^{t _2}dt}$

$=\dfrac{\int _0^t(ma)(at)dt}{\int _0^tdt}$

$=ma^2\dfrac{t}{2}$

$=1\times 4^2\times\dfrac{2}{2}W$

$=16W$

A given $1700 kg$ car goes from $18 {m}/{s}$ to $0 {m}/{s}$. If this transition took $9 sec$, what was the average power supplied by the force causing this deceleration?
Take the car system to be otherwise isolated (i.e. the decelerating force was the only force acting on the car).

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

  1. $-1700 W$

  2. $-15300 W$

  3. $-30600 W$

  4. $-61200 W$

  5. Cannot be determined from the information given

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Correct Option: C
Explanation:

Given :   $u =18$ m/s             $u = 0$ m/s                $m = 1700$ kg             $t = 9$ s

Using   $v = u+at$
$\therefore$  $0 = 18 + a \times 9$                 $\implies a = -2$  $m/s^2$
Using       $v^2  - u^2 = 2aS$
$\therefore$  $0 - (18)^2  =2(-2) S$                      $\implies S = 81$  m

Thus work done by decelerating force        $W  = -ma S $
$\therefore$  Power supplied     $P =\dfrac{-maS}{t}  = \dfrac{- 1700 \times 2 \times 81}{9}  =-30600$  W