Question

P=1/2kx^
If the spring constant is 25600 N/m. How much must thr spring stretch in meters in order to store 8J of potential energy?

Answer 1

Answer:

Extension, x = 0.025m

Explanation:

Given the following data;

Spring constant = 25600 N/m

Potential energy = 8 Joules

To find the extension;

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;

$8 = \frac {1}{2}*25600*x^{2}$

Cross-multiplying, we have;

$16 = 25600x^{2}$

$x^{2} = \frac {16}{25600}$

$x = \sqrt {0.000625}$

Extension, x = 0.025m