Tag: work and power

Questions Related to work and power

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

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

An area of land is an average of $2\ m$ below sea level. To prevent flooding, pumps are used to lift rainwater up to sea level. What is the minimum pump output power required to deal with $1.3 \times 10^9\ Kg$ of rain per day?

  1. $15\ KW$

  2. $30\ KW$

  3. $100\ KW$

  4. $300\ KW$


Correct Option: D
Explanation:

Given,

$m=1.3\times 10^9 kg$
$g=10m/s^2$
$h=2m$
$t=1day$
Power, $P=\dfrac{mgh}{t}$
$P=\dfrac{1.3\times 10^9\times 10\times 2}{1\times 24\times 60\times 60}$
$P=300\times 10^3=300kW$
The correct option is D.

A stone is projected with velocity $u$ at an angle $\theta$ with horizontal. Find out average power of the gravity during time t.

  1. $mg\, \left [ \displaystyle \frac{gt^2}{2}\, - u sin \theta \right ]$

  2. $mg\, \left [ \displaystyle \frac{gt}{2}\, + u sin \theta \right ]$

  3. $mg\, \left [ \displaystyle \frac{gt}{2}\, - u sin \theta \right ]$

  4. $mg\, \left [ \displaystyle \frac{gt}{4}\, - u sin \theta \right ]$


Correct Option: C