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 frictionless 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 1.962 m/s², with block 1 moving downward and block 2 moving upward.

Step-by-step solution

Analyze the forces acting on the blocks

First, we need to determine the gravitational forces acting on both blocks due to their mass, which can be calculated as follows: \(F_{g_1} = m_1 g = 3.00 kg \times 9.81 m/s^2\) \(F_{g_2} = m_2 g = 4.00 kg \times 9.81 m/s^2\) where \(g\) is the acceleration due to gravity and is approximately equal to \(9.81 m/s^2\).

Apply Newton's second law to each block

Now, we can write an equation for each block using Newton's second law, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration: For block 1: \(F_{net_1} = F_{g_1} - T = m_1 a\) For block 2: \(F_{net_2} = T - F_{g_2} = m_2 a\) where \(T\) is the tension in the string and \(a\) is the acceleration of the blocks.

Solve for the acceleration

Since \(F_{net_1} = m_1 a\) and \(F_{net_2} = m_2 a\), we can set up an equation to eliminate the tension \(T\): \((F_{g_1} - T) + (T - F_{g_2}) = m_1 a + m_2 a\) Simplifying the equation gives: \(F_{g_1} - F_{g_2} = (m_1 + m_2) a\) Now, we can plug in the values of \(F_{g_1}\), \(F_{g_2}\), \(m_1\), and \(m_2\) and solve for the acceleration, \(a\): \((3.00 kg \times 9.81 m/s^2) - (4.00 kg \times 9.81 m/s^2) = (3.00 kg + 4.00 kg) \times a\) Calculating the acceleration gives: \(a = \frac{(3.00 kg \times 9.81 m/s^2) - (4.00 kg \times 9.81 m/s^2)}{(3.00 kg + 4.00 kg)}\) \(a = -1.962 m/s^2\) The negative sign indicates that the acceleration of block 1 is downward, and the acceleration of block 2 is upward. The magnitude of the acceleration is \(1.962 m/s^2\).