#### 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