Question

.4 .78 Two blocks of masses \(m_{1}\) and \(m_{2}\) 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. If \(m_{1}=3.50 \mathrm{~kg},\) what value does \(m_{2}\) have to have for the system to experience an acceleration of \(a=0.400 \mathrm{~g}\) ? (Hint: There are two solutions to this problem.)

Short answer

The possible values for mass \(m_2\) are approximately \(9.55 \thinspace kg\) and \(1.45 \thinspace kg\).

Step-by-step solution

Setup the problem using Newton's second law of motion

First, we will label the forces acting on our system. Let \(T\) be the tension in the string, and \(m_1g\) and \(m_2g\) be the gravitational force acting upon the masses \(m_1\) and \(m_2\), respectively. Now we will write down the equations for the individual masses using Newton's second law: For mass \(m_1\): \(T - m_1g = m_1a\) For mass \(m_2\): \(m_2g - T = m_2a\)

Solve the system of equations for \(m_2\)

We want to solve this system of equations for \(m_2\). To do this, we will first eliminate the unknown \(T\). Let's add the two equations to obtain: \((T - m_1g) + (m_2g - T) = m_1a + m_2a\) We can rewrite this as: \((m_2g - m_1g) = (m_1+m_2)a\) Now we are only left with one equation that involves \(m_2\). Let's solve for \(m_2\): \(m_2 = \frac{m_1a + m_1g}{a + g}\)

Substitute known values into the equation

Now substitute the given values for \(m_1 = 3.50 \thinspace kg\), \(a = 0.400g\) and \(g = 9.81 \frac{m}{s^2}\) into the expression for \(m_2\): \(m_2 = \frac{3.50 \thinspace kg(0.400(9.81 \frac{m}{s^2}) + 9.81 \frac{m}{s^2})}{(0.400(9.81 \frac{m}{s^2}) + 9.81 \frac{m}{s^2})}\)

Calculate \(m_2\)

After simplifying and performing the calculations, we can find \(m_2\): \(m_2 = \frac{3.50(3.924 + 9.81)}{3.924 + 9.81} \approx 9.55 \thinspace kg\) Now, as hinted, there is a second solution. We must also consider a scenario in which mass \(m_1\) is moving upward and \(m_2\) is moving downward. Let's revise our original Newton's second law equations for this scenario: For mass \(m_1\): \(m_1g - T = m_1a\) For mass \(m_2\): \(T - m_2g = m_2a\) As in step 2, we will eliminate the unknown \(T\) and solve for \(m_2\). The revised equation is: \((m_1g - T) + (T - m_2g) = m_1a + m_2a\) Now, again we are left with one equation that involves \(m_2\): \((m_1g - m_2g) = (m_1+m_2)a\) Solving for \(m_2\) gives: \(m_2 = \frac{m_1a-m_1g}{a-g}\) Substitute the given values and solve: \(m_2 = \frac{3.50 \thinspace kg(0.400(9.81 \frac{m}{s^2}) - 9.81 \frac{m}{s^2})}{(0.400(9.81 \frac{m}{s^2}) - 9.81 \frac{m}{s^2})} \) \(m_2 = \frac{3.50(3.924 - 9.81)}{3.924 - 9.81} \approx 1.45 \thinspace kg\)

Present the final answers

The possible values for mass \(m_2\) that result in an acceleration of \(0.400g\) for the described Atwood machine setup are \(m_2 \approx 9.55 \thinspace kg\) and \(m_2 \approx 1.45 \thinspace kg\).