Question

When a $65kg$ cheerleader stands on a vertical spring, the spring compresses by $5.5cm$. When a second cheerleader stands on the shoulders of the first, the spring compresses an additional $4.5cm$. What is the mass of the second cheerleader?

Short answer

The mass of second cheerleader is$53kg$.

Step-by-step solution

Content Introduction

Hooke's law says that the force applied on spring is directly proportional to deformation of spring.

$$\begin{gathered} F_{sp}\alpha x \\ {F}_{sp}=kx \end{gathered}$$

Here, $F_{sp}$is force applied on spring, $k$is spring constant, $x$is deformation of spring.

Content Explanation

Let mass of first cheerleader is $m_1$and mass of second cheerleader is $m_2$.

The value of spring constant when force applied by first cheerleader during compression is $5.5cm$

$$\begin{gathered} m_{1}g=kx \\ (65kg)(9.81m/s^2)=k(5.5cm)\left(\frac{1m}{100cm}\right) \\ k=116\times {10}^{2}N/m \end{gathered}$$

Total force applied on spring is

$$\begin{gathered} (m_1+m_2)g=kx \\ (65kg+m_2)(9.81m/s^2)=(116\times {10}^{2}N/m)(10.0cm)\left(\frac{1m}{100cm}\right) \\ (65kg+m_2)=118kg \\ {m}_2=53kg \end{gathered}$$