Answer
The mass of the second cheerleader is 53.2 kg
Work Step by Step
When the first cheerleader stands on the spring, the spring's force is equal to the weight of the cheerleader. We can find the spring constant. Let $m_1$ be the mass of the first cheerleader.
$F_s = m_1~g$
$kx = m_1~g$
$k = \frac{m_1~g}{x}$
$k = \frac{(65~kg)(9.80~m/s^2)}{0.055~m}$
$k = 11,580~N/m$
We can find the mass $m_2$ of the second cheerleader. Note that the spring compression is $10~cm$ when both cheerleaders stand on the spring.
$F_s = (m_1+m_2)~g$
$kx = (m_1+m_2)~g$
$kx-m_1~g = m_2~g$
$m_2 = \frac{kx-m_1~g}{g}$
$m_2 = \frac{(11,580~N/m)(0.10~m)-(65~kg)(9.80~m/s^2)}{9.80~m/s^2}$
$m_2 = 53.2~kg$
The mass of the second cheerleader is 53.2 kg
