Question

A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstretched. The mass is released and falls, stretching the spring by 24 cm before coming to rest at its lowest point. It then continues to oscillate vertically. Part A Part complete What is the spring constant?

Answer 1

Answer:

The spring constant is 45.94 N/m.

Explanation:

Given that,

Length = 50 cm

Mass = 270 g

Stretching the spring = 24 cm

We need to calculate the spring constant

Using formula of energy

The change in potential energy equal to the change in kinetic energy.

$mgh=\dfrac{1}{2}kx^2$

Put the value into the formula

$270\times10^{-3}\times9.8\times50\times10^{-2}=\dfrac{1}{2}\times k\times(24\times10^{-2})^2$

$k=\dfrac{2\times270\times10^{-3}\times9.8\times50\times10^{-2}}{(24\times10^{-2})^2}$

$k=45.94\ N/m$

Hence, The spring constant is 45.94 N/m.