Answer
The 1.50-kg mass should be glued a distance of 9.53 cm to the left of the original center of gravity.
Work Step by Step
Let $m_1$ be the 5.00-kg object and let $m_2$ be the 1.50-kg mass. Let the original center of gravity be the origin. Let $r_2$ be the distance to the left of the original center of mass, where we will glue the 1.50-kg mass.
$x_{cog} = \frac{m_1~r_1+m_2~r_2}{m_1+m_2}$
$m_2~r_2 = x_{cog}~(m_1+m_2)-m_1~r_1$
$r_2 = \frac{x_{cog}~(m_1+m_2)-m_1~r_1}{m_2}$
$r_2 = \frac{(2.20~cm)(5.00~kg+1.50~kg)-(5.00~kg)(0)}{1.50~kg}$
$r_2 = 9.53~cm$
The 1.50-kg mass should be glued a distance of 9.53 cm to the left of the original center of gravity.
