Question

A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the force constant of the spring?What is the unloaded length of the spring?

Answer 1

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

$\Delta F= k \Delta x$

where

$\Delta F = F_2 - F_1$ is the change in the force on the spring, where

$F_1 = m_1 g = (0.300)(9.8)=2.94 N$ is the force applied when the hanging mass is

$m_1 = 0.300 kg$

$F_2 = m_2 g = (1.95)(9.8)=19.1 N$ is the force applied when the hanging mass is

$m_2 = 1.95 kg$

$\Delta x=x_2 -x_1$ is the change in extension of the spring, where

$x_1=0.200 m$ is the extension of the spring when the hanging mass is 0.300 kg

$x_2=0.750 m$ is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

$k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m$

2)

When the first mass is hanging on the spring, we have

$F_1 = k (x_1 - x_0)$

where:

$F_1$ is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

$x_1$ is the extension of the spring wrt its natural length

$x_0$ is the natural length of the spring (the unloaded length)

Here we have

$F_1=2.94 N$

k = 29.5 N/m

$x_1=0.200 m$

Solving for $x_0$, we find:

$x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m$