Asked by falcon306
A spring exerts a 17 N force after being stretched by 3.0 cm from its equilibrium length. By how much will the spring force increase if the spring is stretched from 5.0 cm away from equilibrium to 6.0 rm cm from equilibrium?
I don't know how to approach this at all or how to solve it.
I don't know how to approach this at all or how to solve it.
Answers
Answered by
drwls
The spring constant k is 17/3 = 5.67 N/cm
The force increase going from x = 5 to x = 6 cm deflection is k*(6-5) = 5.67 N
The force increase going from x = 5 to x = 6 cm deflection is k*(6-5) = 5.67 N
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