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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.

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Most popular questions from this chapter

You are at the shoe store to buy a pair of basketball shoes that have the greatest traction on a specific type of hardwood. To determine the coefficient of static friction. \(\mu\), you place each shoe on a plank of the wood and tilt the plank to an angle \(\theta\), at which the shoe just starts to slide. Obtain an expression for \(\mu\) as a function of \(\theta\).

A bosun's chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass \(M=90.0 \mathrm{~kg}\). a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope? b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude \(a=2.0 \mathrm{~m} / \mathrm{s}^{2},\) with what maximum magnitude of force must he null on the rone?

offee filters behave Ince small parachutes, with a drag force that is proportional to the velocity squared, \(F_{\text {drag }}=K v^{2}\). A single coffee filter, when dropped from a height of \(2.0 \mathrm{~m}\), reaches the ground in a time of \(3.0 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag force remains the same, but the weight is doubled. Find the time for the combined filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

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