# An elevator has mass 800 kg, not including passengers. The elevator is designed to ascend, at...

## Question:

An elevator has mass 800 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 17.0 s and it is driven by a motor that can provide up to 37 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has a mass of 68.0 kg.

## Energy Conservation:

The energy conservation principle tells us that all types of energy have been pre-existent. There is no way to make or erase energy. It can also transform from one type to another. This knowledge of energy is essential in harnessing energy to make things easier for the human race. Knowing that energy cannot be erased, we can input variables that will manipulate energy into becoming what we need it to be.

Given:

{eq}m_e = 800 \ kg \\ m_p = 68 \ kg \\ h = 20 \ m \\ t = 17 \ s \\ P = 37 \ hp \times \dfrac {746 \ W}{1 hp} = 27,602 \ W {/eq}

FInd: the number of passengers {eq}x {/eq} that can ride the elevator

Let:

{eq}g = 9.81 \dfrac {m}{s^2} {/eq} , acceleration due to gravity

{eq}v = \dfrac {h}{t} = \dfrac {20 \ m}{17 \ s} = 1.18 \ \dfrac {m}{s} {/eq} , average velocity of the elevator

{eq}m_T = m_e + x m_p {/eq} , the total mass of the system

{eq}E = P \times t = 27,602 \ W \times 17 \ s = 469,234 \ J {/eq} , the total energy the motor can produce one-way

Now, the motor must produce enough to get everyone up, which means:

{eq}E = PE + KE \\ E = m_Tgh+ \dfrac {1}{2} m_T v^2 \\ E = (m_E + xm_P)(gh+ \dfrac {1}{2} v^2) \\ x = \dfrac {\dfrac {E}{gh+ \dfrac {1}{2} v^2} - m_E}{m_P} \\ x = \dfrac {\dfrac {469,234 \ J}{(9.81 \dfrac {m}{s^2})(20 \ m)+ \dfrac {1}{2} (1.18 \dfrac {m}{s})^2} - 800 \ kg}{68 \ kg} \\ x = 23.28 \ passengers = 23 passengers {/eq}

The elevator can handle 23 passengers.