Problem 85 - College Physics

Question

\(\bullet$$$\bullet\) Elevator design. You are designing an elevator for a hos- pital. The force exerted on a passenger by the floor of the ele- vator 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 approximately 5.94 m/s.

Step 1: Define the Problem

Given that the maximum force exerted on the passenger by the floor is 1.60 times their weight, we need to determine the maximum speed the elevator can reach during its upward acceleration. Using the concept of forces, the force exerted can be expressed in terms of the weight and acceleration.

Step 2: Apply Newton's Second Law

Using Newton's Second Law, \( F = ma \), where \( a \) is the acceleration of the elevator, \( F = 1.60 mg \) is the maximum force, and \( m \) is the mass of the passenger. The equation involving the passenger's weight \( mg \) is set as \[ 1.60 mg = m(a + g) \] where \( g \) is the acceleration due to gravity (9.81 m/s²).

Step 3: Solve for Acceleration

Simplify the equation from Step 2 to isolate acceleration: \[ 1.60 mg = m(a + g) \Rightarrow a + g = 1.60g \Rightarrow a = 0.60g = 0.60 \times 9.81 \] Calculate the result to find \( a = 5.886 \text{ m/s}^2 \).

Step 4: Use Kinematic Equation for Maximum Speed

The kinematic equation \( v^2 = u^2 + 2as \) is used to find the maximum speed \( v \) where initial speed \( u \) is 0, acceleration \( a \) is from Step 3, and distance \( s = 3.0 \text{ m} \). Substitute the values \[ v^2 = 0 + 2 \times 5.886 \times 3.0 \].

Step 5: Solve for Maximum Speed

Calculate the maximum speed \( v \) by taking the square root: \[ v^2 = 35.316 \text{ m}^2/ ext{s}^2 \Rightarrow v = \sqrt{35.316} \text{ m/s} \] resulting in \( v \approx 5.94 \text{ m/s} \).

Key Concepts

Newton's Second Law

Newton's Second Law is a cornerstone of classical mechanics, helping us understand how forces affect the motion of an object. The law is expressed with the equation \( F = ma \), where \( F \) is the force applied to an object, \( m \) is its mass, and \( a \) is the acceleration produced. This relationship shows that increasing the force on an object results in greater acceleration. In our elevator design problem, Newton's Second Law helps us relate the force exerted on a passenger to the elevator's acceleration. Here, the challenge is to ensure the force does not exceed 1.60 times the passenger's weight. We set this condition as \( F = 1.60 mg \), where \( g \) is the acceleration due to gravity. Introducing Newton's Second Law, \( F = m(a + g) \), we see how it allows us to solve for the necessary acceleration \( a \). This method provides the basis for understanding how forces affect motion in a controlled environment, like an elevator.

Kinematic Equations

Kinematic equations describe the motion of an object without considering the forces in play. They are especially useful when dealing with constant acceleration, as seen in our elevator problem. The key kinematic equation here is \( v^2 = u^2 + 2as \), where \( v \) is the final velocity, \( u \) the initial velocity, \( a \) the acceleration, and \( s \) the distance traveled.In the context of the exercise, the elevator starts from rest, meaning \( u = 0 \). The acceleration \( a \) has already been derived using Newton's Second Law, and the distance \( s \) is given as 3.0 meters. By substituting these values into the kinematic equation, we can easily compute the maximum speed \( v \) the elevator reaches. Breaking it down to manageable steps makes finding the solution straightforward and demonstrates the power of kinematic equations in motion analysis.

Acceleration Due to Gravity

Acceleration due to gravity, denoted by \( g \), is a constant that affects all objects in free fall near the Earth's surface. It is approximately \( 9.81 \, \text{m/s}^2 \). This constant is vital for calculating forces and motion in physics problems involving vertical movement, such as our elevator scenario.Gravity acts downwards, influencing the effective net force experienced by objects, and is always considered in calculations like in this problem. Here, it combines with the elevator's own acceleration to determine the total force exerted on the passenger. By setting the equation \( 1.60 mg = m(a + g) \), we solve for \( a \), showing the adjustive measure that accounts for both the elevator's acceleration and the force of gravity. Without understanding the role of \( g \), it would be challenging to gauge realistic scenarios involving vertical motion.