Question

An Atwood machine is set up like the figure which is given below.
If $m_1=6.0kg$ and $m_2=4.5kg$. What is the acceleration of the system when the masses are released? (Take the value of $g=10m{s}^{-2}$)

• A. $\frac{10}{7}m{s}^{-2}$
• B. $\frac{20}{7}m{s}^{-2}$
• C. $\frac{10}{3}m{s}^{-2}$
• D. $\frac{20}{3}m{s}^{-2}$

Answer 1

The correct option is A $\frac{10}{7}m{s}^{-2}$

Generally the acceleration of the blocks in the case of Atwood machine is calculated as:
$a=\frac{(m_2-m_1)g}{m_1+m_2}$ As ($m_2>m_1$)
where, $m_1$ and $m_2$ is the mass of block 1 and 2 and $g$ is the acceleration due to gravity.

Here, $m_1=6kg$ and $m_2=4.5kg$
Therefore,
$a=\frac{(4.5-6)\times 10}{4.5+6}=-\frac{10}{7}m{s}^{-2}$