Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 73

A block of mass \(m_{1}=3.00 \mathrm{~kg}\) and a block of mass \(m_{2}=4.00 \mathrm{~kg}\) are suspended by a massless string over a friction less pulley with negligible mass, as in an Atwood machine. The blocks are held motionless and then released. What is the acceleration of the two blocks?

Short Answer

Expert verified
Answer: The acceleration of the two blocks is 1.40 m/s², with block m1 going upwards and block m2 going downwards.

Step by step solution

01

Determine forces acting on the blocks

First, let's identify the forces acting on each block. For block m1, there are two forces: the gravitational force (weight) acting downward, with magnitude \(m_1g\) and the tension force T acting upward. For block m2, there are also two forces: the gravitational force (weight) acting downward, with magnitude \(m_2g\) and the tension force T acting upward. Note that the tension force is the same for both blocks since the string is massless.
02

Apply Newton's second law to each block

Next, we will apply Newton's second law of motion to each block separately. Let's denote the acceleration of block m1 with a1 and the acceleration of block m2 with a2. For block m1: \(m_1a_1 = T - m_1g\) For block m2: \(m_2a_2 = T - m_2g\) Since the string is massless and the pulley is frictionless, the magnitudes of the accelerations of the two blocks are equal: \(a_1 = a_2\). We can denote this common acceleration as a: \(a = a_1 = a_2\).
03

Solve for the tension force T

Now, we have two equations with two unknowns, T and a. Let's solve for T first. From equation of block m1: \(T = m_1a + m_1g\) Substitute this expression for T into the equation of block m2 to eliminate T: \(m_2a = (m_1a + m_1g) - m_2g\)
04

Solve for the acceleration a

Now, we will solve for the acceleration a. Rearrange the equation from step 3: \(a(m_1 + m_2) = m_1g - m_2g\) Now, we can solve for a: \(a = \frac{m_1g - m_2g}{m_1 + m_2}\) Plug in the given values for the masses, \(m_1 = 3.00\,\text{kg}\), and \(m_2 = 4.00\,\text{kg}\), and use \(g = 9.81\,\text{m/s}^2\) for the acceleration due to gravity: \(a = \frac{(3.00\,\text{kg})(9.81\,\text{m/s}^2) - (4.00\,\text{kg})(9.81\,\text{m/s}^2)}{3.00\,\text{kg} + 4.00\,\text{kg}}\)
05

Calculate the acceleration

Finally, calculate the acceleration a: \(a = -\frac{(1.00\,\text{kg})(9.81\,\text{m/s}^2)}{7.00\,\text{kg}}\) \(a = -1.40\,\text{m/s}^2\) The negative sign indicates that the acceleration of block m1 (the lighter block) is in the direction opposite to the positive direction we initially assumed when writing the Newton's second law equations. Therefore, the acceleration of the two blocks is 1.40 m/s², with block m1 going upwards and block m2 going downwards.

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

On the bunny hill at a ski resort, a towrope pulls the skiers up the hill with constant speed of \(1.74 \mathrm{~m} / \mathrm{s}\). The slope of the hill is \(12.4^{\circ}\) with respect to the horizontal. A child is being pulled up the hill. The coefficients of static and kinetic friction between the child's skis and the snow are 0.152 and 0.104 , respectively, and the child's mass is \(62.4 \mathrm{~kg}\), including the clothing and equipment. What is the force with which the towrope has to pull on the child?

A skydiver of mass \(83.7 \mathrm{~kg}\) (including outfit and equipment) falls in the spread-eagle position, having reached terminal speed. Her drag coefficient is \(0.587,\) and her surface area that is exposed to the air stream is \(1.035 \mathrm{~m}\). How long does it take her to fall a vertical distance of \(296.7 \mathrm{~m} ?\) (The density of air is \(1.14 \mathrm{~kg} / \mathrm{m}^{3}\).)

A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to the front of a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. A third mass \(M_{3}=2.50 \mathrm{~kg}\), which is also initially at rest on a frictionless table, is attached to the back of \(M_{2}\) by a light string. a) Find the magnitude of the acceleration, \(a,\) of mass \(M_{3}\) b) Find the tension in the string between masses \(M_{1}\) and \(M_{2}\).

A pinata of mass \(M=12\) kg hangs on a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m}\), the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole, and the total length of the rope between the poles is \(L=3.0 \mathrm{~m}\). The pinata is attached to a ring, with the rope passing through the center of the ring. The ring is frictionless, so that it can slide freely

Leonardo da Vinci discovered that the magnitude of the friction force is usuzlly simply proportional to the magnitude of the normal force; that is, the friction force does not depend on the width or length of the contact area. Thus, the main reason to use wide tires on a race car is that they a) Iook cool. b) have more apparent contact area. c) cost more. d) can be made of softer materials.
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Work Energy and Power

Read Explanation

Space Physics

Read Explanation

Nuclear Physics

Read Explanation

Oscillations

Read Explanation

Electricity

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