Question
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?
a. What is the force constant of this spring?
b. What was the maximum force required to stretch it by that distance?
Answers
drwls
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.
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.