EXERCISE 4.11 Suppose in the same Atwood setup another string is attached to the bottom of m_(1) and a constant force / is applied, retarding the upward motion of m_(1) If m_(1)=5.00kg and m_(1)=10.00kg. what value of fwill reduce the accelera tion of the system by 50% __

To solve this problem, we need to consider the forces acting on the system and how they affect the acceleration. Let's break down the steps:Step 1: Identify the forces in the systemWithout the retarding force, the only forces acting on the system are the gravitational forces on \(m_{1}\) and \(m_{2}\), and the tension in the string. The net force causing the acceleration of the system is the difference in the gravitational forces on \(m_{1}\) and \(m_{2}\), which is \(m_{2}g - m_{1}g\).Step 2: Calculate the initial acceleration of the systemUsing Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the total mass of the system, and \(a\) is the acceleration. The total mass of the system is \(m_{1} + m_{2}\). The initial acceleration \(a_{0}\) is given by:\[a_{0} = \frac{m_{2}g - m_{1}g}{m_{1} + m_{2}}\]Step 3: Introduce the retarding force\(f\) is applied to retard the upward motion of \(m_{1}\). This force will act in the opposite direction to the acceleration, thus reducing the net force and the acceleration. The new net force is \(m_{2}g - m_{1}g - f\), and the new acceleration \(a_{1}\) is:\[a_{1} = \frac{m_{2}g - m_{1}g - f}{m_{1} + m_{2}}\]Step 4: Determine the retarding force needed to reduce acceleration by 50%We want the new acceleration \(a_{1}\) to be half of the initial acceleration \(a_{0}\):\[a_{1} = \frac{1}{2}a_{0}\]Substitute the expression for \(a_{0}\) into the equation for \(a_{1}\) and solve for \(f\):\[\frac{1}{2} \cdot \frac{m_{2}g - m_{1}g}{m_{1} + m_{2}} = \frac{m_{2}g - m_{1}g - f}{m_{1} + m_{2}}\]Step 5: Solve for \(f\)Multiply both sides by \(2(m_{1} + m_{2})\) to clear the fraction:\[m_{2}g - m_{1}g = 2(m_{2}g - m_{1}g - f)\]Distribute the 2 on the right side:\[m_{2}g - m_{1}g = 2m_{2}g - 2m_{1}g - 2f\]Combine like terms:\[m_{2}g - 2m_{2}g = -2m_{1}g + m_{1}g - 2f\]\[-m_{2}g = -m_{1}g - 2f\]Divide by -1 to get:\[m_{2}g = m_{1}g + 2f\]Subtract \(m_{1}g\) from both sides:\[m_{2}g - m_{1}g = 2f\]Factor out \(g\):\[g(m_{2} - m_{1}) = 2f\]Finally, divide by 2 to solve for \(f\):\[f = \frac{g(m_{2} - m_{1})}{2}\]Step 6: Plug in the values for \(m_{1}\) and \(m_{2}\)Given that \(m_{1} = 5.00 \, \text{kg}\) and \(m_{2} = 10.00 \, \text{kg}\), we can plug these values into the equation:\[f = \frac{9.8(m_{2} - m_{1})}{2}\]\[f = \frac{9.8(10.00 - 5.00)}{2}\]\[f = \frac{9.8 \times 5.00}{2}\]\[f = \frac{49.00}{2}\]\[f = 24.50 \, \text{N}\]【Answer】: The value of \(f\) that will reduce the acceleration of the system by 50% is 24.50 N.