Chapter 7: Q7.7-39 PE (page 262)

(a) Find the useful power output of an elevator motor that lifts a $2500 - {\rm{kg}}$ load a height of $35.0{\rm{ m}}$ in $12.0{\rm{ s}}$, if it also increases the speed from rest to $4.00{\rm{ m}}/{\rm{s}}$. Note that the total mass of the counterbalanced system is $10,000{\rm{ kg}} - $so that only $2500{\rm{ kg}}$ is raised in height, but the full $10,000{\rm{ kg}}$ is accelerated.
(b) What does it cost, if electricity is $\$ 0.0900{\rm{ per kW}} \cdot {\rm{h}}$?

Short Answer

Expert verified

(a) The useful power output of an elevator motor is $1.36 \times {10^5}{\rm{ W}}$ .

(b) The cost of operation is $4.08{\rm{ cents}}$ .

Step by step solution

Definition of Concepts

Power: The rate at which the energy of the system changes is known as power.

Mathematically,

$P = \frac{E}{T}$

Here, $E$ is the change in the energy of the system and $T$ is the time.

Find the useful power output of an elevator motor

(a)

The mass of the system is,

${m_s} = {m_e} + {m_c}$

Here,${m_s}$is the mass of the system$\left( {{m_s} = 10000{\rm{ kg}}} \right)$,${m_e}$is the mass of the elevator$\left( {{m_e} = 2500{\rm{ kg}}} \right)$, and${m_c}$is the mass of the counterweight.

The mass of the counterweight is,

${m_c} = {m_s} - {m_e}$

Putting all known values,

$\begin{aligned}{m_c} &= \left( {10000{\rm{ kg}}} \right) - \left( {2500{\rm{ kg}}} \right)\\ &= 7500{\rm{ kg}}\end{aligned}$

The net work done in lifting the elevator is,

$\begin{aligned}W &= \left| {\Delta {\rm{KE}} + \Delta {\rm{PE}}} \right|\\ &= \left| {\frac{1}{2}{m_s}\left( {v_f^2 - v_i^2} \right) + {m_e}gh - {m_c}gh} \right|\end{aligned}$

Here,${m_s}$is the mass of the system$\left( {{m_s} = 10000{\rm{ kg}}} \right)$,${v_f}$is the final velocity of the elevator$\left( {{v_f} = 4.00{\rm{ m}}/{\rm{s}}} \right)$,${v_i}$is the initial velocity of the elevator$\left( {{v_i} = 0} \right)$,${m_e}$is the mass of the elevator$\left( {{m_e} = 2500{\rm{ kg}}} \right)$,$g$is the acceleration due to gravity$\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)$,$h$is the height$\left( {h = 35.0{\rm{ m}}} \right)$, and${m_c}$is the mass of the counterweight$\left( {{m_c} = 7500{\rm{ kg}}} \right)$.

Putting all known values,

$\begin{aligned}W &= \left| \begin{aligned}{l}\frac{1}{2} \times \left( {10000{\rm{ kg}}} \right) \times \left( {{{\left( {4{\rm{ m}}/{\rm{s}}} \right)}^2} - {{\left( 0 \right)}^2}} \right) + \left( {2500{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {35{\rm{ m}}} \right)\\ - \left( {7500{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {35{\rm{ m}}} \right)\end{aligned} \right|\\ &= 1.635 \times {10^6}{\rm{ J}}\end{aligned}$

The useful power output of the elevator is,

$P = \frac{W}{T}$

Here,$T$is the time$\left( {T = 12.0{\rm{ s}}} \right)$.

Putting all known values,

$\begin{aligned}P &= \frac{{1.635 \times {{10}^6}{\rm{ J}}}}{{12.0{\rm{ s}}}}\\ &= 1.36 \times {10^5}{\rm{ W}}\end{aligned}$

Therefore, the useful power output of an elevator motor is $1.36 \times {10^5}{\rm{ W}}$.

Find the cost of operating

(b)

The cost of electricity is,

$c = \$ 0.0900{\rm{ per kW}} \cdot {\rm{h}}$

The cost of electricity bill is,

${\rm{Cost}} = c \times W$

Putting all known values,

$\begin{aligned}{\rm{Cost}} &= \left( {\$ 0.0900{\rm{ per kW}} \cdot {\rm{h}}} \right) \times \left( {1.635 \times {{10}^6}{\rm{ J}}} \right)\\ &= \left( {\$ 0.0900{\rm{ per kW}} \cdot {\rm{h}}} \right) \times \left( {1.635 \times {{10}^6}{\rm{ J}}} \right) \times \left( {\frac{{1{\rm{ kW}} \cdot {\rm{h}}}}{{3.6 \times {{10}^6}{\rm{ J}}}}} \right)\\ &= \$ 0.0408\\ &= 4.08{\rm{ cents}}\end{aligned}$

Therefore, the required cost of operation is $4.08{\rm{ cents}}$.