Question

An elevator that can carry a maximum load of $1800kg$(elevator + passenger) is moving up with a constant speed of$2m{s^ - }^1$. The frictional force opposing the motion is 4000N. What is the minimum power delivered by the motor to the elevator?
(A) $22kW$
(B) $44kW$
(C) $66kW$
(D) $88kW$

Answer 1

Hint: Here the elevator is moving up and frictional force always opposes relative motion. Thus, frictional force will be in downward direction. The force due to load of the elevator will always be in the downward direction.

Complete Step by step solution:
Given, the maximum load the elevator can carry is${M_L} = 1800kg$.
The gravitational force due to this load is given as,
${F_L} = {M_L}g$
$ \Rightarrow {F_L} = 1800 \times 9.8$
Then, we get
$ \Rightarrow {F_L} = 17640N$
The frictional force opposing the motion of the elevator is
$f = 4000N$
The elevator is moving up with a constant velocity of
$v = 2m{s^ - }^1$
We know that frictional force always opposes relative motion. So, when the elevator is moving up the frictional force is in downward direction.
Total force in downward direction is
${F_D} = {F_L} + f$
$ \Rightarrow {F_D} = 17640 + 4000$
So we have
${F_D} = 21640N$
We know that, power delivered to a system moving with constant velocity can be written as
$Power,P = Force \times velocity$
Or, $P = {F_D} \times v$
Putting numeric values into the above equation, we get
$P = 21640 \times 2$
$ \Rightarrow P = 43280W$
Converting the units from $W$to $kW$, we get
$P = \dfrac{{43280}}{{1000}}kW$
$ \Rightarrow P = 43.2kW$
It can be approximated according to the options given in the problem as, we have
$P \simeq 44kW$

Hence, Option (B) is correct.

Note: Here, I have used $g = 9.8m{s^ - }^2$to get an accurate answer. But it is very important to keep in mind that you can take $g = 10m{s^ - }^2$and still get the correct answer in less time. It is preferable to take $g = 10m{s^ - }^2$in exam if the options are rounded off. If the options have decimal values, you should proceed with$g = 9.8m{s^ - }^2$.