Tag: work and energy

Questions Related to work and energy

Calculate the number of Joules in 1KWh.

  1. $\displaystyle 6\times { 10 }^{ -19 }J$

  2. $\displaystyle 3.6\times { 10 }^{ 6 }J$

  3. $\displaystyle 60J$

  4. $\displaystyle 59J$


Correct Option: B
Explanation:

$1$ kW $ = 1000$ $\dfrac{J}{s}$

$1$ h $=3600$ s 
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =3.6\times 10^6$  $J$

When we pay for our electricity bill, we are paying for the ____________.

  1. charge used

  2. current used

  3. power used

  4. energy used


Correct Option: D
Explanation:

A unit is defined as $kWh$, means a $1000 joule/ sec$ is used for $1 hour$ ,

$1KwH=1000\times 3600=3.6\times10^6joules$
we pay for unit and that is energy 

$1kWh= $ _________?

  1. $3600000\ J$

  2. $10000\ J$

  3. $4.2\ J$

  4. $25000\ J$


Correct Option: A
Explanation:
Kilowatt hour is the energy consumed by a body of power $1\ kW$ in $1\ hr$. 
Hence, 
$1\ kWh = 1 kW \times 1\ hr$
              $=  10^3 W \times 3600\ s$
              $= 3600000\ J$

Kilowatt-hour is the unit of :

  1. potential difference.

  2. electric power.

  3. electrical energy.

  4. charge.


Correct Option: C
Explanation:

The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour. The kilowatt-hour is commercially used as a billing unit for energy delivered to consumers by electric utilities.

Kilowatt hour is the unit of:

  1. Power

  2. Energy

  3. Impulse

  4. Force


Correct Option: B
Explanation:

KiloWatt is the unit of power and hour is the unit of time. Product of power and time equals Energy.

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to  :

  1. ${t^{3/4}}$

  2. ${t^{3/2}}$

  3. ${t^{1/4}}$

  4. ${t^{1/2}}$


Correct Option: B
Explanation:

Let's consider a body is moved along a straight line by a machine delivering a constant power $P$. The distance moved by the body is $S$. 


Power, $P=F.v$.. . . . . . (1)


Force, $F=ma$ . . . . . . . .(2)

where, $v=\dfrac{S}{t}$ 

acceleration, $a=\dfrac{S}{t^2}$

$m=$ mass

from  equation (1) and equation (2), we get

$P=\dfrac{mS}{t^2}\times \dfrac{S}{t}$

$S^2=\dfrac{Pt^3}{m}$

From the above equation, we get

$S^2\propto t^3$

$S\propto t^{3/2}$

The correct option is B.

A body is moving along a straight line delivering power given as P = at, then work (W) done is given for time 0 to t is

  1. W = a

  2. W = $\frac{1}{2}a t^2$

  3. W = $2at^2$

  4. W = $a^2t^2$


Correct Option: B
Explanation:

Given that ,

Power , $P= at$
We know that 
$P=\dfrac{dW}{dt}$

$\implies \dfrac{dW}{dt}= at$

$\implies dW= at dt$
$\implies W= \int _0^t atdt$
$\implies W=\dfrac12 at^2$

$\therefore $ Work done for time 0 to t ,   $W=\dfrac12 at^2$  

A force of $ 2\hat { i } +3\hat { j } +2kN $ acts on a body for 4 s and produces a displacement of $3\hat {i} +4\hat {j} +5 \hat {k} m $ calculate the power ?

  1. 5 w

  2. 6 w

  3. 7 W

  4. 9 w


Correct Option: C
Explanation:

Given that,

Force  ,$F= 2 \hat i + 3\hat j + 2\hat k $  N
Displacement , $S= 3\hat i + 4\hat j +5 \hat k $ m 
Time Taken , $t= 4\ s$

Power, $P=\dfrac Pt = \dfrac{F\cdot S}{t}= \dfrac{6+12+10}{4}= \dfrac{28}4 = 7\ W$

A pump of $200W$ power is lifting $2kg$ water from an average depth of $10m$ in one second. Velocity of water delivered by the pump is :

$(g=10m/s^2)$

  1. $10m/s$

  2. $2m/s$

  3. $4 m/s$

  4. $1 m/s$


Correct Option: A
Explanation:

acceleration due to gravity $g = 10m/sec^2$

height $H = 10 meters$
mass $= 2kg$
Potential energy $= mgh = 2\times 10 \times 10 = 200J$
as power $= \dfrac{work \ done}{time}$
when power of motor $= 200w$
$200 = 200/t$
$\Rightarrow t = 1sec$
here displacement of water= height $= 10m$
time = 1sec
Hence,
Velocity $V= \dfrac{Displacement}{Time}$
              $V= \dfrac{10}{1}$
              $V= 10m/sec$

A small diesel engine uses a volume of $1.5 \times 10^4\, cm^3$  of fuel per hour to produce a useful power
output of 40 kW. It may be assumed that 34 kJ of energy is transferred to the engine when it uses $1.0\, cm^3$  of fuel.
What is the rate of transfer from the engine of energy that is wasted?

  1. 850 kW

  2. 920 kW

  3. 840 kW

  4. 810 kW


Correct Option: D
Explanation:

Energy produced by $1.0\ cm^3$ of fuel = $34\ kJ$

So, energy produced by  $1.5 \times 10^4\ cm^3$ of fuel in one hour = $1.5 \times 10^4 \times 34\ kJ$
                                                                                                   = $5.1 \times 10^5\ kJ$
Energy produced in one second = $\dfrac{5.1 \times 10^5}{60}\ kJ/s$
                                                         = $850\ kW$
So, rate of energy wasting = $(850-40)\ kW$
                                             = $810\ kW$