a. Solve the equation
E = 9.50 J
= (1/2) k X^2
for k, using X = 0.020 m
The answer will be in N/m/
b. Force = k X
I don't know why they call that the "maximum" force. You can't stretch it that far with less force.
To stretch a certain spring by 2.00 cm from its equilibrium position requires 9.50 J of work.
a. What is the force constant of this spring?
b. What was the maximum force required to stretch it by that distance?
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