Question

A 50 g mass hanger hangs motionless from a partially stretched spring. When a 80 gram mass is added to the hanger, the spring stretch increases by 8 cm. What is the spring constant of the spring (in N/m)

Answer 1

Answer:

$k = 9.807\,\frac{N}{m}$

Explanation:

The description of both systems can be done by means of the following two equations of equilibrium:

$\Sigma F = k\cdot x - (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )=0$

$\Sigma F = k \cdot (x + 0.08\,m) - (0.13\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = 0$

By diving the first expression by the second one, the following relation is found:

$\frac{x}{x + 0.08\,m}=\frac{0.05\,kg}{0.13\,kg}$

$x = 0.385\cdot (x+0.08\,m)$

$0.615\cdot x = 0.031\,m$

The initial elongation of the spring is:

$x = 0.05\,m$

The spring constant is found by substituting known variable in any of the two equations of equilibrium. For comfort and quickness reasons, the first formula is used:

$k = \frac{(0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.05\,m}$

$k = 9.807\,\frac{N}{m}$