Question

A 2.60-kg mass hangs vertically downward from a massless spring of spring constant 114.0 n/m. What is the length that the spring stretches from its equilibrium position when the mass is hung from the spring?

Answer 1

The spring stretches approximately 0.224 m from its equilibrium position when a 2.60-kg mass is hung from a massless spring with a spring constant of 114.0 N/m.

Step-by-step explanation:

The student has asked for the length the spring stretches from its equilibrium position when a 2.60-kg mass is hung from a massless spring with a spring constant of 114.0 N/m. To solve this, we need to use Hooke's Law which relates the force exerted by the spring and its displacement from equilibrium. The force exerted by the spring (F) is equal to the spring constant (k) times the displacement (x), stated as F = kx. Here, the force is also the gravitational force acting on the mass, which is F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s2).

Setting these two forces equal gives us mg = kx, which we can rearrange to solve for x: x = (mg) / k. Plugging in the values: x = (2.60 kg)(9.8 m/s2) / 114.0 N/m, we find that x is approximately 0.224 m. Therefore, the spring stretches 0.224 m from its equilibrium position when a 2.60-kg mass is hung from it.