Question

A hockey stick stores 4.2 J of potential energy when it is bent 3.1 cm. Treating the hockey stick as a spring, what is its spring constant

Answer 1

Answer:

Spring constant, k = 8740.89 N/m

Explanation:

Given the following data;

Potential energy = 4.2 J

Extension, e = 3.1 cm to meters = 3.1/100 = 0.031 m

To find the spring constant;

Mathematically, the potential energy stored in a spring is given by the formula;

$P.E = \frac {1}{2}Kx^{2}$

Where;

P.E is potential energy.

K is the spring constant.

x is the extension of the spring.

Substituting into the equation, we have;

$4.2 = \frac {1}{2}*k*0.031^{2}$

Cross-multiplying, we have;

$8.4 = k*0.000961$

$k = \frac {8.4}{0.000961}$

Spring constant, k = 8740.89 N/m