Question

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 56 m in 4.9 min, starting and ending at rest. The elevator's counterweight has a mass of only 900 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answer 1

Explanation:

Formula for work and power is as follows.

                W = Fd

and,        P = $\frac{W}{T}$  

So, force exerted by the elevator is equal to equal to mass multiplied by acceleration due to gravity.

             $F_{E} = 1400 \times 9.81$

                        = 13734 N

Now, this force will be counter acted by the force of the counter weight as follows.

             $F_{c} = 900 \times (-9.91)$

                        = -8919

Net force needed to be lifted and therefore, value of net force will be calculated as follows.

          $F_{net}$ = (13734 - 8919) N

                       = 4815 N

Now, calculate the work as follows.

                W = Fd

                     = $4815 N \times 56 m$

                      = 269640 Nm

Now, we will calculate the power needed as follows.

                  P = $\frac{W}{T}$

                     = $\frac{269640 Nm}{4.9 \times 60 sec}$

                     = 917.14 W

Therefore, average power of the force the motor exerts on the cab via the cable is 917.14 W.