Newtons Laws EX 25

The mass is pulled down by the force of gravity, $F_{G}=mg$, and pulled up by the spring force, $F_{s}=kx$. The sum of the forces on the mass is zero because the problem implies that the mass is just hanging from the spring and it is at rest. We choose the coordinate system and draw the free-body diagram as shown below.

$F_{S}-F_{G}=0$

$F_{S}=F_{G}$

$kx=mg$

$x={mg \over k}$

This equation is true when the hanging mass is equal to 200 g or 250 g. We expect that the spring will stretch more. Let $m_{1}=200$ g and $m_{2}=250$ g. Similarly, let $x_{1}=1.5$ cm and $x_{2}$ needs to be determined. We write the equations for both $x_{1}$ and $x_{2}$:

$x_{1}={m_{1}g \over k}$

$x_{2}={m_{2}g \over k}$

Divide the two equations:

${\frac {x_{2}}{x_{1}}}={\frac {m_{2}}{m_{1}}}$

$x_{2}={x_{1}m_{2} \over m_{1}}$

$x_{2}=1.875$ cm