Question

A spring is hanging from the ceiling. A physicist with mass 90 kg jumps up and grabs ahold of the spring. He then bounces up and down without touching the floor. How far down does the spring stretch from its rest length? The spring constant is k=3768 N/m

Answer 1

To determine the stretch of the spring when a physicist hangs from it, we use Hooke's law, F = kx, with the weight of the physicist as the force. Dividing the weight by the spring constant, 3768 N/m, we find that the spring stretches approximately 0.234 meters from its rest length.

Step-by-step explanation:

To find how far down the spring stretches when the physicist jumps and grabs it, we use Hooke's law, which states that the force (F) required to stretch or compress a spring by some distance (x) is proportional to that distance:

F = kx

Where k is the spring constant and x is the extension or compression of the spring from its rest length. The force applied to the spring in this case is the weight of the physicist, which is the product of his mass (m) and the acceleration due to gravity (g), so F = mg. We rearrange Hooke's law to solve for x:

x = F/k

The physicist's weight would be (90 kg)(9.8 m/s²) = 882 N. Using the provided spring constant (k = 3768 N/m), we get:

x = 882 N / 3768 N/m

This calculation yields the stretch of the spring from its rest length. Performing the division, we find that:

x = 0.234 meters

Thus, the spring stretches approximately 0.234 meters from its rest length when the 90 kg physicist grabs onto it.