Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 65

Two masses are joined by a massless string. A \(30-\mathrm{N}\) force applied vertically to the upper mass gives the system a constant upward acceleration of \(3.2 \mathrm{m} / \mathrm{s}^{2} .\) If the string tension is \(18 \mathrm{N},\) what are the two masses?

Short Answer

Expert verified
The mass of the upper object is \( m_{1} = \frac{12N}{3.2 m/s^{2}} \) and the mass of the lower object is \( m_{2} = \frac{18N}{3.2 m/s^{2}} \).

Step by step solution

01

Identify the forces acting on each object

First, let's analyze the upper mass. It experiences two forces: the force applied (30N pointing upward), and the tension in the string (18N downwards). The net force can be calculated by subtracting the force of tension from the applied force. The lower mass only experiences one force, the tension of the string downwards.
02

Setup and calculate the mass of the upper object

Use Newton's Second law of motion (F=ma) to find the mass of the upper object. We know that net force on the upper object is the difference between the applied force and the tension which gives us 30N - 18N = 12N. The acceleration is given as 3.2 m/s^2. To find the mass, we rearrange the equation to m = F/a. So the mass of the upper object is \( m_{1} = \frac{F}{a} = \frac{12N}{3.2 m/s^{2}} \).
03

Setup and calculate the mass of the lower object

Similarly, for the lower mass, the force is the tension in the string which is 18N. The acceleration is the same as the upper mass, which is 3.2 m/s^2. Using the formula m=F/a, the mass of the lower object is \( m_{2} = \frac{F}{a} = \frac{18N}{3.2 m/s^{2}} \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In a front-end collision, a \(1300-\mathrm{kg}\) car with shock-absorbing bumpers can withstand a maximum force of \(65 \mathrm{kN}\) before damage occurs. If the maximum speed for a nondamaging collision is \(10 \mathrm{km} / \mathrm{h},\) by how much must the bumper be able to move relative to the car?

You step into an elevator, and it accelerates to a downward speed of \(9.2 \mathrm{m} / \mathrm{s}\) in \(2.1 \mathrm{s} .\) How does your apparent weight during this time compare with your actual weight?

A \(35-\mathrm{N}\) force is applied to a spring with spring constant \(k=220 \mathrm{N} / \mathrm{m} .\) How much does the spring stretch?

A car leaves the road traveling at \(110 \mathrm{km} / \mathrm{h}\) and hits a tree, coming to a stop in 0.14 s. What average force does a seatbelt exert on a \(60-\mathrm{kg}\) passenger during this collision?

Does a body necessarily move in the direction of the net force acting on it?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Astrophysics

Read Explanation

Kinematics Physics

Read Explanation

Electromagnetism

Read Explanation

Particle Model of Matter

Read Explanation

Thermodynamics

Read Explanation
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.

Sign-up for free