(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 \(\$ 0.0900{\rm{ per kW}} \cdot {\rm{h}}\)?

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.

(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 \(1.36 \times {10^5}{\rm{ W}}\).

(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 \(4.08{\rm{ cents}}\).