Answer
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Hint: Power is defined as the amount of energy converted per unit time or the rate with respect to time at which work is done. The mathematical equation to calculate power is
${\rm{P = \vec F}}{\rm{.\vec V}}$
Where F is force and v is velocity.
Complete step by step answer:
Mass of an elevator $({\rm{m)}} = 920{\rm{kg}}$
The elevator moves with constant speed,${\rm{v}}\;{\rm{ = }}3m/s$
Frictional force opposing the motion ${\rm{f}}\;{\rm{ = 6000N}}$
Also the mass of men ${\rm{M = 68}} \times {\rm{10 = 680 kg}}$
Hence the elevator is moving in an upward direction, so motion is against gravity, so the gravitational force is also retarding the motion. So the total retarding force which opposes the motion is including the weight of the elevator and men.
Hence total retarding force is
${\rm{F}}\;{\rm{ = }}\,{\rm{f + (m + M)g}}$ ……….. (i)
Here $g$ is the acceleration due to gravity given as $\left( {{\rm{g = 10m/}}{{\rm{s}}^2}} \right)$
To compute the total retarding force put all the values in (i) equation we get,
$ \Rightarrow {\rm{F}}\;{\rm{ = }}\,6000{\rm{ + (920 + 680) (10)}}$
$ \Rightarrow \;{\rm{F = 6000}}\;{\rm{ + (1600)(10)}}$
$ \Rightarrow {\rm{F = 16000 + 6000 N}}\, = \;22000{\rm{N}}$ ……… (ii)
The equation to compute the power required to move the elevator with constant speed in the upward direction is
${\rm{P = }}\,{\rm{F}}{\rm{.V}}$ ……..(iii)
Put the values of F and V in equation (iii) we get,
$\Rightarrow {\rm{P}}\,{\rm{ = 22000 }} \times {\rm{ 3}}\;{\rm{ = 66000watt}}$.
Therefore, the power delivered by motor to the elevator ${\rm{ = 66000watt}}$, hence the correct option is (B).
Note:
Power is the scalar quantity. The unit of power, Watt, can be defined as the one joule of work done in one second. Also, the unit of power can also be expressed as watt-second or kilowatt-hour.
${\rm{P = \vec F}}{\rm{.\vec V}}$
Where F is force and v is velocity.
Complete step by step answer:
Mass of an elevator $({\rm{m)}} = 920{\rm{kg}}$
The elevator moves with constant speed,${\rm{v}}\;{\rm{ = }}3m/s$
Frictional force opposing the motion ${\rm{f}}\;{\rm{ = 6000N}}$
Also the mass of men ${\rm{M = 68}} \times {\rm{10 = 680 kg}}$
Hence the elevator is moving in an upward direction, so motion is against gravity, so the gravitational force is also retarding the motion. So the total retarding force which opposes the motion is including the weight of the elevator and men.
Hence total retarding force is
${\rm{F}}\;{\rm{ = }}\,{\rm{f + (m + M)g}}$ ……….. (i)
Here $g$ is the acceleration due to gravity given as $\left( {{\rm{g = 10m/}}{{\rm{s}}^2}} \right)$
To compute the total retarding force put all the values in (i) equation we get,
$ \Rightarrow {\rm{F}}\;{\rm{ = }}\,6000{\rm{ + (920 + 680) (10)}}$
$ \Rightarrow \;{\rm{F = 6000}}\;{\rm{ + (1600)(10)}}$
$ \Rightarrow {\rm{F = 16000 + 6000 N}}\, = \;22000{\rm{N}}$ ……… (ii)
The equation to compute the power required to move the elevator with constant speed in the upward direction is
${\rm{P = }}\,{\rm{F}}{\rm{.V}}$ ……..(iii)
Put the values of F and V in equation (iii) we get,
$\Rightarrow {\rm{P}}\,{\rm{ = 22000 }} \times {\rm{ 3}}\;{\rm{ = 66000watt}}$.
Therefore, the power delivered by motor to the elevator ${\rm{ = 66000watt}}$, hence the correct option is (B).
Note:
Power is the scalar quantity. The unit of power, Watt, can be defined as the one joule of work done in one second. Also, the unit of power can also be expressed as watt-second or kilowatt-hour.
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