you need to show your work, not only the results. **Problem Statement:**3. Two objects of masses M and m (M > m) are connected by a very light flexible string as shown in the figure. Ignoring friction and the mass of the pulley, use Newton’s second law to find the expressions for the acceleration of each block and the tension in the string.**Description of the Diagram:**The diagram shows a pulley system with two masses, M and m, connected by a string. The pulley is mounted on a fixed support. Mass M is hanging on one side of the pulley and mass m is hanging on the other side. ---*Explanation:*To solve for the acceleration and the tension in the string, we start by defining the forces acting on each object.For mass M:- The weight \( W_M \) acts downward: \( W_M = M \cdot g \)- The tension \( T \) in the string acts upward.Using Newton’s second law: \[ M \cdot a = M \cdot g - T \] For mass m:- The weight \( W_m \) acts downward: \( W_m = m \cdot g \)- The tension \( T \) in the string acts upward.Using Newton’s second law: \[ m \cdot a = T - m \cdot g \]By adding the two equations, we eliminate T:\[ M \cdot g - T + T - m \cdot g = M \cdot a + m \cdot a \]\[ (M - m) \cdot g = (M + m) \cdot a \]\[ a = \frac{(M - m) \cdot g}{M + m} \]Next, substitute the expression for acceleration \( a \) back into one of the original equations to solve for the tension \( T \):\[ T = m(g + a) = m \left(g + \frac{(M - m) \cdot g}{M + m}\right) \]\[ T = m \left(\frac{g(M + M - m)}{M + m}\right) \]\[ T = m \left(\frac{(M + m) \cdot g}{M + m}\right) = \frac{2mgM}{M+m}\]Thus, 1. The acceleration of each block is \( a = \frac{(

Name: you need to show your work, not only the results. **Problem Statement:
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: you need to show your work, not only the results. **Problem Statement:**3. Two objects of masses M and m (M > m) are connected by a very light flexible string as shown in the figure. Ignoring friction and the mass of the pulley, use Newton’s second l