Respuesta :
Answer:
To find the total energy delivered by the motor when the elevator is halfway up and moving at its cruising speed, we need to consider the kinetic energy (KE) and potential energy (PE) of the elevator at that point.
1. Calculate the kinetic energy at cruising speed:
\[
KE = \frac{1}{2} \times \text{mass} \times \text{velocity}^2
\]
\[
KE = \frac{1}{2} \times 1200 \text{ kg} \times (5.5 \text{ m/s})^2
\]
\[
KE = \frac{1}{2} \times 1200 \times 30.25
\]
\[
KE = \frac{1}{2} \times 36300
\]
\[
KE = 18150 \text{ J}
\]
2. Calculate the potential energy at halfway up (42 m above the ground):
\[
PE = \text{mass} \times \text{gravity} \times \text{height}
\]
\[
PE = 1200 \text{ kg} \times 9.81 \text{ m/s}^2 \times 42 \text{ m}
\]
\[
PE = 1200 \times 9.81 \times 42
\]
\[
PE = 494064 \text{ J}
\]
3. The total energy delivered by the motor is the sum of kinetic and potential energy at that point:
\[
\text{Total energy} = KE + PE
\]
\[
\text{Total energy} = 18150 + 494064
\]
\[
\text{Total energy} = 512214 \text{ J}
\]
Therefore, when the elevator is halfway up and moving at its cruising speed, the total energy delivered by the motor is 512,214 J.
