The energy of a spring compressed by 2 cm is 5 mJ. What is the stiffness of the spring?
| June 24, 2021 | education
Given:
dx = 2 centimeters = 0.02 meters – the amount of compression of the spring;
W = 5 mJ = 0.005 Joules – potential energy of a compressed spring.
It is required to determine k (N / m) – spring stiffness.
To determine the stiffness of the spring, use the following formula:
W = k * dx ^ 2/2, hence:
k = 2 * W / dx ^ 2 = 2 * 0.005 / 0.02 ^ 2 = 0.01 / 0.0004 = 25 Newtons / meter.
Answer: the coefficient of stiffness of the spring is 25 N / m
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