/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 You are riding in an elevator on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are riding in an elevator on the way to the 18th floor of your dormitory. The elevator is accelerating upward with \(a =\) 1.90 m/s\(^2\). Beside you is the box containing your new computer; the box and its contents have a total mass of 36.0 kg. While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_k =\) 0.32, what magnitude of force must you apply?

Short Answer

Expert verified
You must apply 135 N of force.

Step by step solution

01

Determine the Normal Force

When the elevator accelerates upwards, the apparent weight of the box increases due to the acceleration. To find the normal force, we start by calculating the apparent weight using the formula: \[ \text{Normal Force (N)} = m(g + a) \]where:- \( m = 36.0 \text{ kg} \) is the mass of the box,- \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity,- \( a = 1.9 \text{ m/s}^2 \) is the upward acceleration of the elevator.Substituting the values, we have:\[ N = 36.0 \times (9.8 + 1.9) = 36.0 \times 11.7 = 421.2 \text{ N} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's laws of motion
Newton's laws of motion form the foundation of classical mechanics. The second law, in particular, explains how the motion of an object is influenced by the forces acting upon it. In this exercise, when the elevator accelerates upward, the force applied on the box includes not only gravity, but also the effect of that acceleration. According to Newton's second law, the sum of these forces results in a change in motion.
  • The second law can be expressed as: \[F = ma\]where \( F \) is the total force, \( m \) is the mass, and \( a \) is the acceleration.
  • This equation tells us that the acceleration of an object is directly dependent on the net force acting on it and inversely dependent on its mass.
Since the elevator is moving upward, two forces add up: gravitational force (weight) and the force due to the elevator's acceleration. In this case, the normal force supporting the box is greater than just the weight because of the added upward acceleration. Understanding these laws allows us to predict how objects will move in various conditions.
Kinetic friction
Kinetic friction is the resisting force between two surfaces sliding past one another. In our scenario, the box of the computer and the elevator floor create this type of friction as one is sliding over the other. It's always important to consider this friction because it influences the effort required to move the object.The frictional force that must be overcome to slide the box is calculated using:\[F_k = \mu_k \cdot N\]where:
  • \( F_k \) is the force of kinetic friction,
  • \( \mu_k \) is the coefficient of kinetic friction (given as 0.32 in this problem),
  • \( N \) is the normal force determined in the previous section.
By applying a horizontal force that matches \( F_k \), the box moves at a constant speed. Since it moves steadily, not accelerating, the applied force must be equal and opposite to the kinetic friction force. This balance is crucial in ensuring that the box slides smoothly.
Apparent weight
Apparent weight is the perception of weight under the influence of acceleration. It changes when an object is in a moving or accelerating system, like an elevator. Actually, it's not the weight of the object changing, but rather the force it exerts on the surface.When the elevator is at rest, the apparent weight equals the true weight calculated by:\[W = mg\]However, with the upward acceleration, the formula adjusts to:\[W_{apparent} = m(g + a)\]A box in an accelerating elevator feels heavier because the elevator adds to the pull of gravity. This exponential force enhances what we perceive as weight. Understanding this concept helps clarify why the normal force increases when the elevator moves up. This rising normal force correlates with the box's increased apparent weight.
Normal force
Normal force is the support force exerted perpendicular to the surface that an object contacts. It's what keeps the object from "falling through" the surface. It acts in response to gravity plus any additional forces, such as an elevator accelerating upwards.In the context of the elevator ride problem, calculating normal force is essential because it directly affects the frictional force required to move the box.
  • The formula for normal force when an object rides an accelerating elevator is:\[N = m(g + a)\]where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( a \) is the acceleration from the elevator.
  • This calculation results in a substantial force, due to the added effect of the elevator's motion.
The significance of normal force lies in its role in determining the kinetic friction. A greater normal force means more force needed to move an object, as seen with the box in the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 50.0-kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is 95.0 m/s, what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00\(g\)? (b) What is the apparent weight of the pilot at the lowest point of the pullout?

A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t) =\) (3.0 m/s\(^2\))\(t\) \(+\) (0.20 m/s\(^3\))\(t^2\). When \(t =\) 4.0 s, what is the reading on the bathroom scale?

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

A 2.00-kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. At \(t =\) 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude \(F(t) = (6.00 N/s^2)t^2\). (a) What distance does the box move from its position at \(t =\) 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at \(t =\) 3.00 s?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.