/*! 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 73 Two boxes are at rest on a smoot... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a force of magnitude \(F=12.00 \mathrm{N}\), the contact force between the boxes is \(8.50 \mathrm{N} ;\) if, instead, box 2 is pushed with the force \(F\) the contact force is \(12.00 \mathrm{N}-8.50 \mathrm{N}=3.50 \mathrm{N}\). In either case, the boxes move together with an acceleration of \(1.70 \mathrm{m} / \mathrm{s}^{2}\). What is the mass of (a) box 1 and (b) box \(2 ?\)

Short Answer

Expert verified
Box 1 has mass 2.06 kg; Box 2 has mass 5.00 kg.

Step by step solution

01

Understand the Problem

We are given two scenarios to analyze two boxes in contact on a smooth surface with forces applied. We must find the individual masses of the boxes using the forces and acceleration provided.
02

Identify Concepts Involved

The problem involves concepts of Newton's second law of motion which states that \( F = ma \), where \( F \) is the net force applied just to a single box, \( m \) is the mass, and \( a \) is the acceleration.
03

Develop Equations for Each Scenario

1. When box 1 is pushed, the force\( F = 12.00 \text{ N} \), and the contact force is \(8.50 \text{ N}\). Therefore: \[ F - F_{contact} = m_1a \], where \( F_{contact} = 8.50 \text{ N} \) and \( a = 1.70 \text{ m/s}^2 \). Substituting gives: \[ 12.00 - 8.50 = m_1 \times 1.70 \].2. When box 2 is pushed, the force is \( F \) again, and the contact force is \(3.50 \text{ N}\), so \[ F_{contact} + m_1a = F \]. Using \( F_{contact} = 3.50 \text{ N} \) \[ 3.50 + m_1 \times 1.70 = 12.00 \].
04

Solve for Box 1's Mass

From the equation obtained: \[ 12.00 - 8.50 = m_1 \times 1.70 \]\[ 3.50 = m_1 \times 1.70 \]\[ m_1 = \frac{3.50}{1.70} \approx 2.06 \text{ kg} \].
05

Solve for Box 2's Mass

Calculate box 2's mass using the acceleration calculated:Using Newton's second law:\[ m_2 \times 1.70 = 8.50 \text{ N} + 2.06 \times 1.70 \]\[ m_2 \times 1.70 = 8.50 \text{ N} + 3.50 \text{ N} \]\[ m_2 \times 1.70 = 12.00 \text{ N} \] \[ m_2 = \frac{12.00}{1.70} - m_1 \]\[ m_2 \approx 7.06 - 2.06 \]\[ m_2 = 5.00 \text{ kg} \] .
06

Verify the Solution

Verify using both equations to ensure consistency:1. Calculations for \( m_1 = 2.06 \text{ kg} \) were consistent.2. Calculations for \( m_2 = 5.00 \text{ kg} \) also matched condition from both pushing scenarios using net forces.

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.

Force and Motion
When it comes to understanding how objects move, Newton's Second Law plays a crucial role in describing their behavior. The law is expressed with the formula: \[ F = ma \] where: - \( F \) is the force applied to an object. - \( m \) is the object's mass. - \( a \) is the acceleration that the object experiences. This essential law helps us connect the dots between force and motion. In our example problem, we deal with two boxes, and forces are applied to them, causing movement. It’s by understanding this movement and interaction of forces that we can determine the mass of each box. When pushing Box 1 with a force of 12.00 N and seeing the resulting contact force of 8.50 N, or pushing Box 2 and observing a different contact force, we're witnessing direct applications of Newton’s principles. These changes in motion tell us about the individual masses of the boxes, allowing us to calculate the forces each experiences relative to their mass and the shared acceleration.
Mass Calculation
Calculating mass from the forces and motions observed involves deducing how much mass each object in the question has, given the force dynamics observed. From Newton's Second Law, we rearrange to find mass as:\[ m = \frac{F}{a} \] In our problem, the two connected boxes have different forces working on them depending on where the force \( F \) is applied. Let's break this down: 1. **Box 1 Calculation**: When Box 1 is pushed: - \[ F - F_{contact} = m_1 imes a \] - By substituting the given contact force, acceleration, and total pushing force, we determine the mass of Box 1. - This translates to: \[ m_1 = \frac{12.00 \text{ N} - 8.50 \text{ N}}{1.70 \text{ m/s}^2} = 2.06 \text{ kg} \] 2. **Box 2 Calculation**: When Box 2 is pushed: - \[ m_2 \times a + m_1 \times a = F \] - Using details from both scenarios - we know Box 2’s mass by the result of the combined forces when it is pushed. By clearly understanding these equations and carefully substituting values, we learn precise masses of both boxes in contact.
Smooth Surface Dynamics
A crucial aspect of solving this type of problem is recognizing that the physics of motion changes depending on the surface conditions. A "smooth surface" suggests minimal friction, which simplifies our analysis by focusing on direct force and motion correlation without friction interfering. - **Low Friction Impact**: With such a surface, the motion of the boxes is only influenced by the applied and contact forces, not slowed by the friction typically found in real-world interactions. - **Simplified Calculations**: This removal of friction means our calculations for force and motion become more straightforward. We directly apply Newton's Second Law without worrying about an extra frictional force to subtract. In essence, understanding that the problem takes place on a smooth horizontal surface ensures that all the forces applied are making their full expected impact on the acceleration, leading to more pure and clear results when unraveling the mystery of mass through motion.

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 \(9.3-\mathrm{kg}\) child sits in a \(3.7-\mathrm{kg}\) high chair. (a) Draw a freebody diagram for the child, and find the normal force exerted by the chair on the child. (b) Draw a free-body diagram for the chair, and find the normal force exerted by the floor on the chair.

A farm tractor tows a \(3700-\mathrm{kg}\) trailer up an \(18^{\circ}\) incline with a steady speed of \(3.2 \mathrm{m} / \mathrm{s} .\) What force does the tractor exert on the trailer? (Ignore friction.)

IP While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 35 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance required is about \(1.5 \mathrm{km}\). (a) If the mass of a jet is \(1.70 \times 10^{5} \mathrm{kg}\) what force is needed for takeoff? (b) Describe the strategy you used to solve part (a).

" IP A drag racer crosses the finish line doing \(202 \mathrm{mi} / \mathrm{h}\) and promptly deploys her drag chute (the small parachute used for braking). (a) What force must the drag chute exert on the \(891-\mathrm{kg}\) ear to slow it to \(45.0 \mathrm{mi} / \mathrm{h}\) in a distance of \(185 \mathrm{m} ?\) (b) Describe the strategy you used to solve part (a).

A \(0.53-\mathrm{kg}\) billiard ball initially at rest is given a speed of \(12 \mathrm{m} / \mathrm{s}\) during a time interval of \(4.0 \mathrm{ms}\). What average force acted on the ball during this time?

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.