Question

What is the speed of the elevator's motor with which it can lift a $1000kg$ load which produces $3000W$?

Answer 1

- Hint: - We know that,
$P = \dfrac{w}{t}$
By which we can express power as-
$P = Fv$
The force of a motor can be calculated by- $F = mg$ (where, $g = 10m{s^{ - 2}}$)
Now, we can find the velocity of a motor by using the force and power of the motor.

Complete Step by Step Solution: -
The rate of work done with respect to time is called Power. We can also say that, it is the ratio of work done and time. It is denoted by $P.$ The S.I unit of power is Watt $(W).$ Mathematically, power can be represented as –
Then, let the work done be $w$ and time be $t.$
$P = \dfrac{w}{t} \cdots (1)$
We know that work done on an object is the product of force applied on an object and displacement of an object.
It can be expressed as –
$w = Fd$
(where, $F$ is the force of an object and $d$ is the displacement of an object)
Putting the value of work done, $w = Fd$ in the equation $(1)$
$P = F\dfrac{d}{t}$
Now, we know that velocity is the ratio of displacement and time taken by an object. So, we get
$P = Fv$ $\left| \!{{\,
  {\because v = \dfrac{d}{t}} \,}} \right. $

$\because F = ma$
$\therefore P = mav \cdots (1)$
Now, according to the question, it is given that
Mass, $m = 1000kg$
Power, $P = 3000W$
In the case of an elevator, the acceleration produced by the elevator's motor is equal to acceleration due to gravity.
$\therefore a = g = 10m{s^{ - 2}}$
Now, putting these values in equation $(1)$, we get
$3000 = 1000 \times 10 \times v$
Now, by transposition, we get
$
  v = \dfrac{{3000}}{{10000}} \\
  v = 0.3m/\sec \\
 $
So, the velocity produced by the motor is $0.3m/\sec $.

Note: - Power is the scalar quantity. The unit of power, Watt can be defined as the one joule of work done in one second. Sometimes, the unit of power can also be expressed as watt – second or kilowatt – hour.