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If a mass of 0.55kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium, what is the spring constant?
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Answers ( )
Weight G=m×g=0.55×9.8N
G=k×dx, where dx=2cm is the stretch.
dx=2cm=0.02m
m×g=k×dx
k=m×g/dx=0.55×9.8/0.02=269.5N/m
Answer: k=269.5N/m