Question

A mass of 0.368 kilograms is suspended from a spring, and the spring stretches 12.3 centimeters. The mass is then removed, and 0.399 joules of work is used to pull the spring down. How far does the spring stretch? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

The gravitational force acting on the mass can be given as,

$F=mg$

The spring constant of the spring can be given as,

$k=\frac{F}{d}$

Substitute the known expression,

$k=\frac{mg}{d}$

Substitute the known values,

$\begin{gathered} k=\frac{(0.368kg)(9.8m/s^2)}{(12.3\text{ cm)(}\frac{1\text{ m}}{100\text{ cm}})}(\frac{1\text{ N}}{1kgm/s^2}) \\ \approx29.3\text{ N/m} \end{gathered}$

The work done on the spring can be given as,

$W=\frac{1}{2}kx^2$

Substitute the known values,

$\begin{gathered} 0.399J=\frac{1}{2}(29.3N/m)x^2 \\ x^2=\frac{2(0.399\text{ J)}}{29.3\text{ N/m}}(\frac{1\text{ Nm}}{1\text{ J}}) \\ x=\sqrt[]{0.0272m^2} \\ \approx0.165\text{ m} \end{gathered}$

Thus, the distance of spring stretched is 0.165 m.