A fully loaded, slow-moving freight elevator has a cab with a total mass of1200 kg, which is required to travel upward 54 m in 3.0 min , starting and ending at rest. The elevator’s counterweight has a mass of only 950 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

Mass of the cab of freight elevator is, ${\mathrm{m}}_{\mathrm{e}}=1200\mathrm{kg}$

Mass of the counter weight of the elevator is, ${\mathrm{m}}_{\mathrm{c}}=950\mathrm{kg}$

Distance required to travel by the elevator is,$\mathrm{d}=54\mathrm{m}$

Time interval required to travel a distance by the elevator is,

$$\begin{gathered} ∆\mathrm{t}=3\mathrm{m} \\ =180\mathrm{s} \end{gathered}$$

This problem is based on the relation between the force and work done. Work done by the force is the product of displacement of an object and the applied force on it in the direction of displacement. To initiate, identify alltheforces involved in this process, and find their corresponding work done. Then, find the total work done . By using these values,the average power prequired of the force the motor exerts on the cab via the cable can be calculated.

Formulae are as follow:

Work done by the gravity is,

$\mathrm{W}=-\mathrm{mgd}$

where, m is the mass, g is an acceleration due to gravity and d is the displacement.

The power is given by,

$\mathrm{P}=\frac{{\mathrm{W}}_{\mathrm{m}}}{∆\mathrm{t}}$

Here, the loaded elevator is moving upwards with a constant speed. Therefore, three forces are involved:

Gravitational force exerted on the elevator and the corresponding work done is $W_e$.

Gravitational force exerted on the counter weight and the corresponding work done is $W_c$.

Force by the motor via cable and the corresponding work done is $W_m$

Therefore, the total work done is given as,

$W=W_e+W_c+W_m$(ii)

Since, the loaded elevator is moving upwards with a constant speed, the kinetic energy must be constant, and hence, the work done due to kinetic energy is zero.

$$\begin{gathered} W=K \\ =0 \end{gathered}$$

Using notation in equation (i), work done by gravity on the elevator is given by,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{e}}=-{\mathrm{m}}_{\mathrm{e}}\mathrm{gd} \\ {\mathrm{W}}_{\mathrm{e}}=-\left(1200\times 9.8\times 54\right) \\ {\mathrm{W}}_{\mathrm{e}}=-635040\mathrm{J} \end{gathered}$$

(iii)

Similarly,using notation in equation (i), work done by the gravity on counter weight is given by,

$$\begin{gathered} {\mathrm{W}}_c=-{\mathrm{m}}_c\mathrm{gd} \\ {\mathrm{W}}_c=-\left(950\times 9.8\times 54\right) \\ {\mathrm{W}}_{\mathrm{e}}=-502740\mathrm{J} \end{gathered}$$

(iv)

And using notation in equation (i), work done by the motor via cable is given by,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{m}}=-{\mathrm{W}}_{\mathrm{e}}-{\mathrm{W}}_{\mathrm{c}} \\ {\mathrm{W}}_{\mathrm{m}}=635040-502740 \\ {\mathrm{W}}_{\mathrm{m}}=1.323\times {10}^5\mathrm{J} \end{gathered}$$

(v)

Time interval for ${\mathrm{W}}_{\mathrm{m}}$is

$$\begin{gathered} ∆\mathrm{t}=3.0\mathrm{m} \\ =180\mathrm{s} \\ {\mathrm{W}}_{\mathrm{m}}=\mathrm{P}∆\mathrm{t} \end{gathered}$$

Therefore, the power supplied by the motor to lift the elevator in interval is,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{W}}_{\mathrm{m}}}{∆\mathrm{t}} \\ \mathrm{P}=\frac{1.323\times {10}^5}{180} \\ \mathrm{P}=735\mathrm{W} \\ \approx 7.4\times {10}^2\mathrm{W} \end{gathered}$$

Hence, the average power required of the force the motor exerts on the cab via the cable is $\mathrm{P}=7.4\times {10}^2\mathrm{W}$