Question

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of $6.0kg$, accelerates downward at $\frac{3}{4}g$. What is the mass of the other block?

Short answer

The mass of the second block is$\frac{6}{7}kg$.

Step-by-step solution

Given information

The rope is a massless, frictionless, and stationary pulley.

$$\begin{gathered} m_1=6.0kg \\ a=\frac{3}{4}g \end{gathered}$$

Explanation

Apply Newton's second law on m₁ and $m_2$with former going down and later going up.

$$\begin{gathered} m_{1}a=m_{1}g-T\dots \dots \left(1\right) \\ {m}_{2}a=T-m_{2}g\dots \dots \left(2\right) \end{gathered}$$

On adding equation $\left(1\right)$and equation $\left(2\right)$

$$\begin{gathered} g\left[m_1-m_2\right]=a\left[m_1+m_2\right] \\ {m}_2=m_1\frac{\left[g-a\right]}{\left[g+a\right]}\dots \dots \left(3\right) \end{gathered}$$

Substituting the given data in the equation $\left(3\right)$

localid="1649613361939" $$\begin{gathered} m_2=\left(6kg\right)\times \frac{\left[g-{\displaystyle \frac{3}{4}}g\right]}{\left[g+{\displaystyle \frac{3}{4}}g\right]} \\ {m}_2=\frac{6}{7}kg \end{gathered}$$

The mass of the second block is$\frac{6}{7}kg$.