Hookes Law, Finding A Springs Constant

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kcrawshaw
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A spring extends by 10cm when a mass of 100g is attached to it. What is the spring constant? (calculate your answer in N/m)

I have done this so far but I don't feel that it is right as the Force (F) is in grams and not Newtons:

F = K x E
100 = K x 0.10
100 ÷ 0.10 = 100
K = 100 N/m

Is this right or have I gone wrong somewhere?
Thanks
 
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What's the weight (in Newtons) of a 100 g mass? (Always convert everything to standard units.)
 
[itex]F- kx[/itex]
100g = 100*10 = 1000 Newtons of force.

[itex]1000 = -k (-0.1) \implies k = \boxed{10 000}[/itex]
 
You'll need to convert the weight from g to kg; right now you have the units in (g*m/s^2)/m
N/m is (kg*m/s^2))/m where the unit of Newtons is kg*m/s^2
 
oneplusone said:
[itex]F- kx[/itex]
100g = 100*10 = 1000 Newtons of force.

[itex]1000 = -k (-0.1) \implies k = \boxed{10 000}[/itex]
This is incorrect.

Please allow the OP to solve the problem for themselves.