Posted by **PHYSICIS HELP ASAP** on Wednesday, November 10, 2010 at 9:31am.

a 0.1 kg mass is suspended at rest from a spring near the Earth's surface, the distance that the spring is stretched is measured to be 1.0 cm. What is the spring constant of the spring (remember the MKS units)?

A mass of 2 kg is attached to a spring with constant 18 N/m. How far is the spring stretched?

A spring with a spring constant of 28.9 N/m is attached to different masses, and is stretched 5cm. What is the force exerted on the spring?

- PHYSICS -
**drwls**, Wednesday, November 10, 2010 at 11:36am
Part 1: A M=0.1 kg mass weighs F = Mg = 0.98 Newtons. The spring constant is

k = F/X . If you put X in meters (0.01 m), do the computation and you will get k in MKS units of N/m

Part 2:

Use the same Hooke's law formula,

k = F/X

Rearrange it to get X= F/k

The do the calculation. They told you what k is. You will have to convert the 2 kh mass to weight force.

Part 3: F = k X

In this case they tell you that X = 0.05 m

Do the numbers.

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