Question

A spring with k = 54 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 30 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

When a 30 kg mass is attached to the spring and is at its lowest position, the spring stretches 5.44 m due to the weight of the mass. Therefore, the end of the spring will line up at approximately 561 cm or 5.61 m on the ruler.

Step-by-step explanation:

To determine where the end of the spring will line up with the ruler marks when a 30 kg mass is attached at its lowest position, we use Hooke's law, which relates the force exerted by a spring to its displacement (F = kx). The force exerted by the mass on the spring is the weight of the mass (mg), where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s2).

The force exerted by the mass is therefore F = (30 kg)(9.8 m/s2) = 294 N. Since F = kx, we can solve for the displacement x: x = F/k. Plugging in the values, we get x = 294 N / 54 N/m.

Calculating this gives us x ≈ 5.44 m or 544 cm. This is the amount the spring stretches due to the weight of the mass. Therefore, the end of the spring will line up with the 17 cm mark on the ruler plus the 544 cm of stretching, giving us a total of 561 cm or 5.61 m from the top of the ruler.