Q80P - University Physics with Modern Physics

Question

You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger’s weight. The elevator accelerates upward with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed of the elevator?

Step-by-Step Solution

Answer

The maximum speed of the elevator is 5.93 m/s .

Step 1: Identification of the given data

The given data can be listed below as:

• The distance moved by the elevator is h = 3.0 m.
• The exerted force is less than   times the weight of the passenger.

Step 2: Significance of the maximum speed

The maximum speed is described as the highest velocity attained by the athlete. The speed is directly proportional to the acceleration of an object.

Step 3: Determination of the maximum speed

According to the question, the equation of the acceleration of the elevator is expressed as:

 

1.6mg - mg = ma

0.6mg = ma

a = 0.6 g  

 

Here, m is the mass and a is the acceleration of the elevator and g is the acceleration due to gravity.

 

Substitute 9.8 $\mathrm{m}/{\mathrm{s}}^2$ for g in the above equation.

 

 $$\begin{gathered} \mathrm{a}=0.6\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =5.88\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

The equation of the maximum speed is expressed as:

 

$\begin{array}{l}{\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{ah}\\ \mathrm{v}=\sqrt{{\mathrm{u}}^2+2\mathrm{ah}}\end{array}$

Here, v is the maximum speed and u is the initial speed. h is the distance moved by the elevator.

 

As the elevator was at rest initially, the initial velocity of the elevator is zero.

 

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{v}=\sqrt{0+2\left(5.88\mathrm{m}/{\mathrm{s}}^2\right)\left(3.0\mathrm{m}\right)} \\ =\sqrt{\left(11.76\mathrm{m}/{\mathrm{s}}^2\right)\left(3.0\mathrm{m}\right)} \\ =\sqrt{\left(35.28{\mathrm{m}}^2/{\mathrm{s}}^2\right)} \\ =5.93 \mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the maximum speed of the elevator is 5.93 m/s .