Question

A spring is stretched from 0.25 meter to a length of 1.50 meters by a weight of 3.20 N. What is the spring contant? Step by step please

Answer 1

2.56 N/m

Step-by-step explanation:

The force exerted by a spring is given by the equation:

`F_s = kx`

where
`k` is the spring constant, and
`x` is the distance that the spring is stretched or compressed from its equilibrium.

To find
`x`, we will simply subtract the spring's initial position from its final position:

`x = 1.50 m - 0.25 m\\x = 1.25 m`

It is given in the problem that the force exerted by the spring is 3.20 Newtons. So, we can solve the equation for
`k` and calculate the spring constant.

`k = (F_s)/(x) \\k = (3.20N)/(1.25m) \\k = 2.56 N/m`