Question

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

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

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.

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: \(a = a_1 = a_2\).

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: \(m_2a = (m_1a + m_1g) - m_2g\)

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: \(a = \frac{(3.00\,\text{kg})(9.81\,\text{m/s}^2) - (4.00\,\text{kg})(9.81\,\text{m/s}^2)}{3.00\,\text{kg} + 4.00\,\text{kg}}\)

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.

Key concepts

Newton's Second Law

Newton's Second Law is a cornerstone of classical mechanics, formulated by Sir Isaac Newton, and tells us that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This relationship is succinctly expressed with the formula: \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
In the case of the Atwood machine, Newton's Second Law helps us understand how the different masses of the two blocks affect their motion. Even though both blocks experience the same gravitational force, their different masses mean they also experience different net forces when released from rest. This ultimately results in different accelerations if the masses were not balanced.
However, since the problem states that the string is massless and we assume there's no friction on the pulley, both blocks actually share the same acceleration value due to the constraints of being connected by the string.

Tension Force

Tension Force is the force exerted by a rope or a string when it is pulled tight by forces acting from opposite ends. In our scenario, the rope or string connecting the two blocks in the Atwood machine transmits the force from one block to the other, enabling them to exert force upon each other.
In an ideal Atwood machine, where the string and pulley have negligible mass and there’s no friction, the tension force in the string is uniform throughout. This means that both blocks experience the same magnitude of tension force despite their different masses. The tension works against the gravitational force on the heavier block, trying to accelerate it upward, while it assists the gravitational force on the lighter block, attempting to pull it downward.

Acceleration Calculation

Once the system is released from rest, calculating the acceleration involves understanding the interplay of forces acting on both blocks. The formula derived from Newton’s Second Law, taking both masses into account, is \( a = \frac{m_1g - m_2g}{m_1 + m_2} \).
This formula is obtained by equating the net forces experienced by the two blocks, since they share the same tension force in the string and their accelerations are equal though opposite in direction. By solving this equation, we are able to determine the linear acceleration of the system once the values for \( m_1 \), \( m_2 \), and \( g \) (acceleration due to gravity) are plugged in.
Hence, specific to this scenario with given masses, the resulting acceleration is calculated to be \( 1.40 \, \text{m/s}^2 \), with the direction being dictated by which block is heavier.

Gravitational Force

Gravitational Force, also known as weight, is the force by which a planet or other celestial body attracts objects toward its center. For objects on or near the Earth's surface, this force is the object's mass multiplied by the Earth's gravitational acceleration, \( g \), which is approximately \( 9.81 \, \text{m/s}^2 \).
Within an Atwood machine, gravitational force acts downward on both blocks. The magnitude of the gravitational force on each block is given by \( m_1g \) and \( m_2g \), respectively, where \( m_1 \) and \( m_2 \) represent the masses of the blocks.
Understanding gravitational force is essential for calculating the resultant forces and, consequently, the acceleration of the blocks, aiding us in predicting their motion once the system is set free.