Tag: energy and its forms

Questions Related to energy and its forms

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. $15.42\times 10^5 W$

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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. $1.5 W$

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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

Show answer
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

Your uncle pushes a $60.0 kg$ crate along a floor with average speed $v=0.65 {m}/{s}$ for $5.0$ seconds as he moves furniture to clean up the garage.
If the coefficient of friction between the floor and the crate is $\mu=0.340$, what is the average power output of your uncle during this time?

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

  1. $26.0 W$

  2. $130 W$

  3. $383 W$

  4. $650 W$

  5. None of the above

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Correct Option: B
Explanation:

Given :  $\mu = 0.340$             $v = 0.65$  m/s                $m =60.0$  kg

As the crate moves with constant speed, thus the force applied by uncles must be equal to the frictional force.
$\therefore$   $F  = \mu mg = 0.340 \times 60.0 \times 9.8  = 199.92$  N
Average power output        $P = F v = 199.92 \times 0.65  \approx 130$  W

A $0.25 kg$ book falls from a height of $1.00 m$, initially at rest. What is the AVERAGE power output of gravity on the book in this time period?

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

  1. $2.45 W$

  2. $5.43 W$

  3. $10.9 W$

  4. $24.1 W$

  5. None of the above

Show answer
Correct Option: B
Explanation:

Given :   $m = 0.25$ kg               $S =1.00$ m                    $u =0$ m/s                $a = g = 9.8m/s^2$

Using   $S = ut+\dfrac{1}{2}at^2$
$\therefore$    $1.00 = 0 + \dfrac{1}{2} \times 9.8t^2$                                     $\implies t = 0.45$ s
 Work done by gravity         $W = mgh  =0.25 \times 9.8 \times 1.00 = 2.45$  J
$\therefore$Output power  of gravity        $P = \dfrac{W}{t} = \dfrac{2.45}{0.45} = 5.43$  $W$

Find the power of a pump which takes $10 s$ to draw $100 kg$ of water from a tank situated at a height of $20 m$.

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

  1. $2\times { 10 }^{ 4 }W$

  2. $2\times { 10 }^{ 3 }W$

  3. $200 W$

  4. $1 kW$

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Correct Option: B
Explanation:

Power = $\dfrac{work done}{time}$


             =  $\dfrac{mgh}{time}$

            =  $\dfrac{100 \ast 10 \ast 20}{10}$

             =  $\dfrac{20000}{10}$

               = 2000W  i.e  2 $\times$ 10$^{3}$W

A $40 N$ force accelerates a $20 kg$ mass through a distance of $16 m$. If there were no other forces acting on the mass and the mass started at rest, what was the average power output of the force?

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

  1. $0 W$

  2. $160 W$

  3. $226 W$

  4. $640 W$

  5. Cannot be determined from the information given

Show answer
Correct Option: B
Explanation:

Given :  $F = 40$ N            $m = 20$ kg              $S = 16$ m               $u = 0$ m/s

Thus acceleration of the mass        $a = \dfrac{F}{m} = \dfrac{40}{20} =2$  $m/s^2$
Using    $S = ut + \dfrac{1}{2}at^2$
$\therefore$  $16 = 0 + \dfrac{1}{2} \times 2 t^2$                      $\implies t = 4$ s
Work done by the force     $W = FS = 40 \times 16  =640$ J
$\therefore$ Power output        $P = \dfrac{W}{t} = \dfrac{640}{4} = 160$  W

An engine of $4.9 kW$ power is used to pump water from a well which is $20 m$ deep. What quantity of water in kiloliters can it pump out in $30$ minutes?

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

  1. $45 kl$

  2. $75 kl$

  3. $25 kl$

  4. $90 kl$

Show answer
Correct Option: A
Explanation:

Power$=\dfrac{work}{time}=\dfrac{mgh}{t}$

$=\dfrac{4.9\times 10^3\times 30\times 60}{9.8\times 20}=m$
$m=45kl$