Tag: work and power

Questions Related to work and power

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

physics work and energy commercial unit of energy power work and power

  1. charge used

  2. current used

  3. power used

  4. energy used

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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= $ _________?

physics work and energy commercial unit of energy power work and power

  1. $3600000\ J$

  2. $10000\ J$

  3. $4.2\ J$

  4. $25000\ J$

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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 :

physics work and energy commercial unit of energy power work and power

  1. potential difference.

  2. electric power.

  3. electrical energy.

  4. charge.

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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:

physics work and energy commercial unit of energy power work and power

  1. Power

  2. Energy

  3. Impulse

  4. Force

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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  :

physics work and power commercial unit of energy power work and energy

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

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

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

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

Show answer
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

physics work and power commercial unit of energy power work and energy

  1. W = a

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

  3. W = $2at^2$

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

Show answer
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 ?

physics work and power commercial unit of energy power work and energy

  1. 5 w

  2. 6 w

  3. 7 W

  4. 9 w

Show answer
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 :

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

physics work and power commercial unit of energy power work and energy

  1. $10m/s$

  2. $2m/s$

  3. $4 m/s$

  4. $1 m/s$

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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?

physics work and power commercial unit of energy power work and energy

  1. 850 kW

  2. 920 kW

  3. 840 kW

  4. 810 kW

Show answer
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 = $(850-40)\ kW$
                                             = $810\ kW$

A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest :

physics work and power commercial unit of energy power work and energy

  1. at the highest position of the body

  2. at the instant just before the body hits the earth

  3. it remains constant throughout

  4. at the instant just after the body is projected

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Correct Option: B
Explanation:

$Power,$ $P$ $=$$\overrightarrow{F}$.$\overrightarrow{v}$ $=$ $Fv$ $cos$$\theta$
$Just$ $before$ $hitting$ $the$ $earth$ $θ$ $=$ $0°.$ $Hence,$ $the$ $power$ $exerted$ $by$ $the$ $gravitational$ $force$ $is$ $greatest$ $at$ $the$ $instant$ $just$ $before$ $the$ $body$ $hits$ $the$ $earth.$