Question

Atwood machine question: details are in photos attached

Answer 1

First, let's calculate the forces in the 3 kg mass:

`W_1-T=m_1\cdot a`

Then, the forces in the 2 kg mass:

`T-W_2=m_2\cdot a`

From the first equation, let's solve it for T:

`T=W_1-m_1a`

Using this value of T in the second equation, we have:

`\begin{gathered} W_1-m_1a-W_2=m_2\cdot a \\ W_1-W_2=m_2a+m_1a \\ m_1g-m_2g=a\cdot(m_1+m_2)_{} \\ a=g((m_1-m_2))/((m_1+m_2)) \end{gathered}`

Finally, using the given values (with g = 10 m/s²), we have:

`a=10\cdot((3-2))/((3+2))=10\cdot(1)/(5)=2\text{ m/s2}`

If the acceleration is 2 m/s², after 2 seconds the speed is:

`\begin{gathered} V=V_0+a\cdot t \\ V=0+2\cdot2 \\ V=4\text{ m/s} \end{gathered}`

Therefore the correct option is the second one.