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Question: An elevator of mass 500kg is to be lifted up at a constant velocity of \(0.4m{{s}^{-1}}\) . What sho...

An elevator of mass 500kg is to be lifted up at a constant velocity of 0.4ms−10.4m{{s}^{-1}} . What should be the minimum horse power of the motor to be used? (Take g=10ms−2g=10m{{s}^{-2}} and 1hp = 750watts)

Explanation

Solution

Since, we need to find out the minimum power, we will use the formula of Power, that is work done by a force per unit time. Now in the problem we have been given the value of velocity. So, we will calculate the power required by using the value of velocity as the distance travelled by the lift in unit time. This will give us both distance and time to be used in our equation.

Complete step-by-step answer:
Let the power required to lift the elevator be given by P.
Also, the force against which this power shall act is the force of gravity, which is equal to:
=mg=mg
Here,
⇒m=\Rightarrow m= mass of elevator which is equal to 500kg
⇒g=10ms−2\Rightarrow g=10m{{s}^{-2}}
=(500×10)N =1000N \begin{aligned} & =(500\times 10)N \\\ & =1000N \\\ \end{aligned}
Now, the velocity of the elevator is constant and given as 0.4ms−10.4m{{s}^{-1}} . This can be broken down into displacement and time as, 0.4 meters is the distance travelled by the elevator in 1 second.
Therefore,
The distance (d) is equal to =0.4m=0.4m
And, the time (t) is equal to =1s=1s
Now, the power delivered to the elevator can be written as:
⇒P=mgdt\Rightarrow P=\dfrac{mgd}{t}
Putting the values of all the known terms in right hand side of the equation, we get:
⇒P=5000×0.41Watt ⇒P=2000Watt \begin{aligned} & \Rightarrow P=\dfrac{5000\times 0.4}{1}Watt \\\ & \Rightarrow P=2000Watt \\\ \end{aligned}
Now, we need to convert this into horse power.
Given:
⇒750Watt=1hp\Rightarrow 750Watt=1hp
Therefore, 2000 Watts will be equal to:
⇒P=2000750 ⇒P=83hp ∴P=2.66hp \begin{aligned} & \Rightarrow P=\dfrac{2000}{750} \\\ & \Rightarrow P=\dfrac{8}{3}hp \\\ & \therefore P=2.66hp \\\ \end{aligned}
Hence, the minimum output of the motor to move the elevator with a uniform speed of 0.4ms−10.4m{{s}^{-1}} should be 2.66 Horse-power.

Note: In this problem, the specific data on distance and time wasn’t provided. So, we made use with the data in hand that was velocity and used it to calculate the power of the motor. Also, one should not confuse between power and work done as both have almost similar formulas. One should take care of these little points, so that we do not concur any error in our solution.