Question

A spring hangs from the ceiling with an unstretched length of x 0 = 0.69 m x0=0.69 m . A m 1 = 7.5 kg m1=7.5 kg block is hung from the spring, causing the spring to stretch to a length x 1 = 0.84 m x1=0.84 m . Find the length x 2 x2 of the spring when a m 2 = 2.1 kg m2=2.1 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.

Answer 1

Answer:

x2=0.732m

Explanation:

We can calculate the spring constant using the equilibrium equation of the block m1. Since the spring is in equilibrium, we can say that the acceleration of the block is equal to zero. So, its equilibrium equation is:

$m_1g-k\Delta x_1=0\\\\\implies k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(7.5kg)(9.8m/s^{2})}{0.84m-0.69m}=490N/m$

Then using the equilibrium equation of the block m2, we have:

$m_2g-k\Delta x_2=0\\\\\\implies x_2=x_0+\frac{m_2g}{k} \\x_2=0.69m+\frac{(2.1kg)(9.8m/s^{2})}{490N/m}= 0.732m$

In words, the lenght x2 of the spring when the m2 block is hung from it, is 0.732m.