Chapter 9, Problem 82PQ

(a)

To determine

The average power of each elevator’s motor during the acceleration.

Answer to Problem 82PQ

The average power of each elevator’s motor during the acceleration is $3.88\times {10}^4\text{W}$ .

Explanation of Solution

Write the expression for the distance travelled during acceleration.

  $\Delta y=\overline{v}t$                                                                                                                        (I)

Here, $\Delta y$ is the distance travelled by elevator, $\overline{v}$ is the average velocity and $t$ is the time taken to reach maximum velocity.

Write the expression for average velocity.

  $\overline{v}=\frac{v_i+v_f}{2}$

Here, $v_i$ is the initial velocity and $v_f$ is the final velocity.

Substitute $\frac{v_i+v_f}{2}$ for $\overline{v}$ in equation (I) to get $\Delta y$.

  $\Delta y=\left(\frac{v_i+v_f}{2}\right)t$                                                                                                       (II)

The forces acting on the elevator car are gravitational force and the force applied by elevator motor.

According to work-energy theorem, net work done by gravitational force and elevator motor is equal to change in kinetic energy.

Write expression for the change in kinetic energy.

  $\Delta K=W_{\text{motor}}+W_g$                                                                                                    (III)

Here, $W_{\text{motor}}$ is the work done by motor and $W_g$ is he work done by gravity and $\Delta K$ is the change in kinetic energy.

Write the expression for the work done by gravity.

  $W_g=mg\Delta y\cos \theta$

Here, $m$ is the mass of elevator full of passengers, $g$ is the acceleration due to gravity and $\theta$ is the angle between gravitational force and displacement of elevator.

Since gravitational force acts in downward direction and elevator moves in upward direction, angle $\theta =180°$.

Substitute $180°$ for $\theta$ in above equation to get $W_g$.

  $\begin{array}{c}{W}_g=mg\Delta y\cos 180°\\ =-mg\Delta y\end{array}$

Write the expression for change in kinetic energy of elevator.

  $\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

Initially elevator is at rest. Substitute $0\text{m}/\text{s}$ for $v_i$ in above equation to get $\Delta K$.

  $\Delta K=\frac{1}{2}m{v}_f^2$

Substitute $\frac{1}{2}m{v}_f^2$ for $\Delta K$, $-mg\Delta y$ for $W_g$ in equation (III) to get $W_{\text{motor}}$.

  $\begin{array}{c}\frac{1}{2}m{v}_f^2=W_{\text{motor}}+-mg\Delta y\\ W_{\text{motor}}=\frac{1}{2}m{v}_f^2+mg\Delta y\end{array}$                                                                                         (IV)

Write the expression for the power of the motor.

  $\overline{P}=\frac{W_{\text{motor}}}{\Delta t}$                                                                                                                (V)

Here, $\overline{P}$ is the average power of elevator motor during acceleration and $\Delta t$ is the time taken to do work.

Conclusion:

It is given that mass of elevator full of passenger is $1155\text{kg}$, final velocity of elevator is $6.10\text{m}/\text{s}$ and time taken to reach this velocity is $5.00\text{s}$.

Substitute $0\text{m}/\text{s}$ for $v_i$, $6.10\text{m}/\text{s}$ for $v_f$ and $5.00\text{s}$ for $t$ in equation (II) to get $\Delta y$.

  $\begin{array}{c}\Delta y=\left(\frac{0\text{m}/\text{s}+6.10\text{m}/\text{s}}{2}\right)\left(5.00\text{s}\right)\\ =15.25\text{m}\end{array}$

Substitute $1155\text{kg}$ for $m$, $6.10\text{m}/\text{s}$ for $v_f$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $15.25\text{m}$ for $\Delta y$ in equation (IV) to get $W_{\text{motor}}$.

  $\begin{array}{c}{W}_{\text{motor}}=\frac{1}{2}\left(1155\text{kg}\right){\left(6.10\text{m}/\text{s}\right)}^2+\left(1155\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(15.25\text{m}\right)\\ =1.94\times {10}^5\text{J}\end{array}$

Substitute $1.94\times {10}^5\text{J}$ for $W_{\text{power}}$ and $5.00\text{s}$ for $\Delta t$ in equation (V) to get $\overline{P}$.

  $\begin{array}{c}\overline{P}=\frac{1.94\times {10}^5\text{J}}{5.00\text{s}}\\ =3.88\times {10}^4\text{W}\end{array}$

Therefore, the average power of each elevator’s motor during the acceleration is $3.88\times {10}^4\text{W}$ .

(b)

To determine

The average power of each elevator’s motor during the cruising phase of its motion.

Answer to Problem 82PQ

The average power of each elevator’s motor during the cruising phase of its motion is $6.88\times {10}^4\text{W}$ .

Explanation of Solution

The elevator attained a cruising speed of $6.10\text{m}/\text{s}$. After this velocity net force on elevator is zero. That is applied force balances weight of the elevator.

Write the expression for the average power during cruising motion.

  $P=Fv$                                                                                                                    (VI)

Here, $P$ is the average power of the motor during cruising motion of elevator , $F$ is the applied force and $v$ is the velocity of elevator.

Write the expression for $F$.

  $F=mg$

Conclusion:

Substitute $1155\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in above equation to get $F$.

  $\begin{array}{c}F=\left(1150\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =11281.5\text{N}\end{array}$

Substitute $11281.5\text{N}$ for $F$ and $6.10\text{m}/\text{s}$ for $v$ in equation (VI) to get $P$.

$\begin{array}{c}P=\left(11281.5\text{N}\right)\left(6.10\text{m}/\text{s}\right)\\ =6.88\times {10}^4\text{W}\end{array}$

Therefore, The average power of each elevator’s motor during the cruising phase of its motion is $6.88\times {10}^4\text{W}$ .