Question

1. the masses in an atwood machine are 1.5 kg and 2.5 kg (see the figure). if the masses are at a height of 1.25 m when the system is released

a) how many seconds does it take for the greater mass to reach the ground?
b) what is the velocity of the smaller mass when the greater one strikes the ground?
c) if this experiment were carried out on the moon, how many seconds would it take for the 2.5 kg mass to fall? (g_e = 10 n/kg·g_m = g_e/6)

Explanation:

Step1: Calculate the acceleration of the system

According to Newton's second law, for the Atwood machine, \(F = ma\). The net force \(F=(m_2 - m_1)g\) and the total mass \(m = m_1 + m_2\). So the acceleration \(a=\frac{(m_2 - m_1)g}{m_1 + m_2}\).
Substituting \(m_1 = 1.5\space kg\), \(m_2 = 2.5\space kg\) and \(g = 10\space N/kg\) into the formula:

$$a=\frac{(2.5 - 1.5)\times10}{1.5 + 2.5}=\frac{1\times10}{4}= 2.5\space m/s^{2}$$

Step2: Use the kinematic equation \(h=v_0t+\frac{1}{2}at^{2}\) to find the time \(t\) for part (a)

Since the initial velocity \(v_0 = 0\space m/s\), the equation simplifies to \(h=\frac{1}{2}at^{2}\).
We know \(h = 1.25\space m\) and \(a = 2.5\space m/s^{2}\). Rearranging for \(t\):

$$t=\sqrt{\frac{2h}{a}}=\sqrt{\frac{2\times1.25}{2.5}}=\sqrt{1}=1\space s$$

Step3: Use the kinematic equation \(v = v_0+at\) to find the velocity \(v\) for part (b)

Since \(v_0 = 0\space m/s\), \(a = 2.5\space m/s^{2}\) and \(t = 1\space s\)

$$v=0 + 2.5\times1=2.5\space m/s$$

Step4: Recalculate the acceleration for the Moon's gravity for part (c)

Using \(g_M=\frac{g_E}{6}=\frac{10}{6}\space N/kg\) in the acceleration formula \(a_M=\frac{(m_2 - m_1)g_M}{m_1 + m_2}\)

$$a_M=\frac{(2.5 - 1.5)\times\frac{10}{6}}{1.5 + 2.5}=\frac{1\times\frac{10}{6}}{4}=\frac{10}{24}\space m/s^{2}$$

Then using \(h=\frac{1}{2}a_Mt_M^{2}\) and solving for \(t_M\)

$$t_M=\sqrt{\frac{2h}{a_M}}=\sqrt{\frac{2\times1.25}{\frac{10}{24}}}=\sqrt{\frac{2.5\times24}{10}}=\sqrt{6}\approx 2.45\space s$$

Answer:

a) \(1\space s\)
b) \(2.5\space m/s\)
c) \(\sqrt{6}\approx 2.45\space s\)