Tag: work, energy and power

Questions Related to work, energy and power

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 $t=2s$.

  1. $16 W$

  2. $1.6 W$

  3. $15 W$

  4. $1.5 W$


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).

  1. $-1700 W$

  2. $-15300 W$

  3. $-30600 W$

  4. $-61200 W$

  5. Cannot be determined from the information given


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?

  1. $26.0 W$

  2. $130 W$

  3. $383 W$

  4. $650 W$

  5. None of the above


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?

  1. $2.45 W$

  2. $5.43 W$

  3. $10.9 W$

  4. $24.1 W$

  5. None of the above


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 $20 m$.

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

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

  3. $200 W$

  4. $1 kW$


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?

  1. $0 W$

  2. $160 W$

  3. $226 W$

  4. $640 W$

  5. Cannot be determined from the information given


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?

  1. $45 kl$

  2. $75 kl$

  3. $25 kl$

  4. $90 kl$


Correct Option: A
Explanation:

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

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

A girl of mass $40 kg$ climbs $50$ stairs each of average height $20 cm$ in $50 s$. Find the power of the girl $\left( g=10m{ s }^{ -2 } \right) $.

  1. $50 W$

  2. $50\times 20W$

  3. $80 W$

  4. $50\times 20\times 2W$


Correct Option: C
Explanation:

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

$=\dfrac { change\quad in\quad PE }{ time } \ =\dfrac { 40\times 10\times 50\times 20\times { 10 }^{ -2 } }{ 50 } \ =80W$

In a factory, due to a sudden strike the work usually done in a day took a longer time. Which of the following happened?

  1. Power increased

  2. Power decreased

  3. Energy increased

  4. None of these


Correct Option: B
Explanation:

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

Since time is increased,power decreased.

In each heart beat, a heart pumps $80\ ml$ of blood at an average pressure of $100\ mm$ of $Hg$. Assuming $60$ heart beats per second, the power output of the heart is $(\rho _{Hg} = 13.6\times 10^{3} kgm^{-3})(g = 9.8\ ms^{-2})$.

  1. $60.97\ W$

  2. $63.97\ W$

  3. $1.12\ W$

  4. $83.97\ W$


Correct Option: B
Explanation:

Power=$P\times\Delta V\times beat\ rate$

$\rho gh\Delta V\times 60$
$13.6\times10^3\times9.8\times 100\times 10^{-3}\times80\times10^{-6}\times60=63.97\ W$