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?

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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.