Question

The question is in the photos (dangling separate weights by two separate ropes question)

Answer 1

In order to calculate the acceleration of the system, let's consider both masses as one object, and use the Second Law of Newton (considering that the blocks are moving down):

`\begin{gathered} \sum ^{}_{}F=m\cdot a \\ W-T=(m_1+m_2)\cdot a \\ (m_1+m_2)\cdot g-T=(m_1+m_2)\cdot a \\ 0.54\cdot9.8-6.6=0.54\cdot a \\ 5.29-6.6=0.54a \\ -1.31=0.54a \\ a=-(1.31)/(0.54)=-2.426\text{ m/s2} \end{gathered}`

So the acceleration is 2.426 m/s² upwards.

The tension in the lower rope can be calculated using the Second law just in the lower block:

`\begin{gathered} \sum ^{}_{}F=m\cdot a \\ T-W_2=m_2\cdot a \\ T-0.24\cdot9.8=0.24\cdot2.426 \\ T-2.352=0.582 \\ T=2.934\text{ N} \end{gathered}`

The tension in the lower rope is equal to 2.934 N.