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\(\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?

Short Answer

Expert verified
The maximum speed of the elevator is approximately 5.94 m/s.

Step by step solution

01

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.
02

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).
03

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 \( a = 5.886 \text{ m/s}^2 \).
04

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 \[ v^2 = 0 + 2 \times 5.886 \times 3.0 \].
05

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 \( v \approx 5.94 \text{ m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

With its wheels locked, a van slides down a hill inclined at \(40.0^{\circ}\) to the horizontal. Find the acceleration of this van a) if the hill is icy and frictionless, and b) if the coefficient of kinetic friction is \(0.20 .\)

What is the acceleration of a raindrop that has reached half of its terminal velocity? Give your answer in terms of \(g .\)

\(\bullet$$\bullet\) Stopping distance of a car. (a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by lock- ing the brakes when traveling at 29.1 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} )\) ? (b) On wet pavement, the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in ordel to be able to stop in the same distance as in part (a)? (Note. Locking the brakes is not the safest way to stop.)

\(\bullet$$\bullet\) You are working for a shipping company. Your job is to stand at the bottom of an 8.0 -m-long ramp that is inclined at \(37^{\circ}\) above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is \(\mu_{k}=0.30\) (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you?

\(\bullet\) A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal sur- face. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a hor- izontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

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