Friday

April 18, 2014

April 18, 2014

Posted by **:kkk:** on Saturday, August 18, 2007 at 3:47am.

equilibrium length of spring: 10.5 cm

stretched length of spring (by attaching a mass): 12.5 cm

object's mass: 198.9 grams

-please help me...I'm really lost...

You need to assume that the measurement was made at a place where the accleration of gravity = g. On the surface of the Earth, that value is 9.8 m/s^2. The force stretching the spring is them

F = m g = 0.1989 kg * 9.8 m/s^2 = ? Newtons

To get the spring constant k, divide F by the strech amount: 0.020 m (which is 20 cm)

The dimensions of your answer will be N/m

k = m g / (change in length)

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