Question

What is the power of the force required to move a4500kgelevator cab with a load of 1800 kgupward at constant speed 3.80 m/s?

Short answer

The power of the force to move an 4500 kg elevator cab with a load of 1800 kg upward at a constant speed 3.80 m/s is 235 KW

Step-by-step solution

Given information

It is given that,

i) Mass of elevator cab 4500 kg

ii) Mass of load 1800 KG

iii) Speed v=3.80 m/s

Determining the concept

Using the formula of power in terms of force and velocity, find the power of force when it moves an object with a constant velocity.

Formula is as follow:

$p=Fv$

Where,

F is force, vis velocity and Pis the power.

Determining the power of the force

Now,

$p=Fv$

But,the total force is,

$$\begin{gathered} F_{total}=\left(\mathrm{mass}\mathrm{of}\mathrm{elevator}\mathrm{cab}+\mathrm{mass}\mathrm{of}\mathrm{load}\right)9.8\mathrm{m}/{\mathrm{s}}^2 \\ {F}_{total}=\left(4500\mathrm{kg}+1800\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{F}}_{\mathrm{total}}=61740\mathrm{N} \end{gathered}$$

Therefore,

$$\begin{gathered} P=\left(61740\mathrm{N}\right)\left(3.80\frac{\mathrm{m}}{\mathrm{s}}\right) \\ P=234612\mathrm{W} \\ =235\mathrm{kW} \end{gathered}$$

Therefore, thepower of the force to move an 4500kg elevator cab with a load of 1800 kg upward at a constant speed 3.80 m/s is 235kW