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

Answer:

Explanation:

The force occurring on a spring F = mg

If x should be the expansion in the spring.

Then F = mg = -kx

where;

$x = \dfrac{mg}{k}$

$x = \dfrac{2.5 \ kg*9.81 \ m/s^2 }{58 \ N/m}$

x = 0.423 m

x = 42.3 cm

The final reading of the spring when the end of the spring lines up with the ruler mark is = 42.3 cm + 17 cm

= 59.3 cm

≅ 60 cm