How much (in m) will a spring that has a force constant of 44.5 N/m be stretched by an object with mass 0.490 kg when hung motionless from the spring?

Respuesta :

Answer:

0.108 m

Explanation:

From Hooke's law, we have that Force applied to a spring and the corresponding extension is given as:

F = ke

where k = spring constant

We also know that weight (force) is given as:

F = mg

where m = mass and g = acceleration due to gravity

=> ke = mg

=> e = mg / k

Therefore, the extension on the spring is:

e = (0.49 * 9.80) / 44.5 = 0.108 m

It will be stretched by 0.108 m