Answer to Question #86974 in Mechanics | Relativity for Maria

Question

A spring with k = 58 N/m hangs vertically next to a ruler. The end of the spring is next to the 17-cm mark on the ruler.
If a 2.5-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?
Express your answer to two significant figures and include the appropriate units.

Answer 1

Let's first find the stretch of the spring from the Hooke's Law:

$$F = kx,$$

here, $F = mg$ is applied force (the force of gravity), $k$ is the spring constant and $x$ is stretch of the spring.

Then, we get:

$$x = \dfrac{F}{k} = \dfrac{mg}{k} = \dfrac{2.5 kg \cdot 9.8 \dfrac{m}{s^2}}{58 \dfrac{N}{m}} = 0.42 m = 42 cm.$$

To find the answer, we need to add up the elongation and the initial position of the end of the spring:

$$x_{end} = x + x_0 = 42 cm + 17 cm = 59 cm.$$

Answer:

$x_{end} = 59 cm.$