Question

An ideal spring obeys Hooke’s law, F~ = −k~x. A mass of 0.50 kilogram hung vertically from this spring stretches the spring 0.075 meter. The acceleration of gravity is 10 m/s 2 . The value of the force constant k for the spring is most nearly

Answer 1

The spring constant is 66.7 N/m

Explanation:

First of all, we have to find the magnitude of the force acting on the spring. This is equal to the weight of the mass hanging on the spring, which is:

$F=mg$

where:

m = 0.50 kg is the mass of the object

$g=10 m/s^2$ is the acceleration of gravity

Substituting,

$F=(0.50)(10)=5 N$

Now we can use Hookes' law to find the constant of the spring:

$F=-kx$

where

F is the force applied

k is the spring constant

x is the stretching of the spring

Here we have:

F = 5 N

While the stretching is

x = 0.075 m

Therefore, ignoring the negative sign in the formula (which only tells us the direction), we find the spring constant:

$k=\frac{F}{x}=\frac{5}{0.075}=66.7 N/m$

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