When a spring is compressed 2.50 × 10^–2 meter

from its equilibrium position, the total potential

energy stored in the spring is 1.25 × 10^–2 joule.

Calculate the spring constant of the spring.

[Show all work, including the equation and

substitution with units.]

The working of a spring is given by:

\(T_{el}= \frac{k\Delta x^2}{2}\)

Entering the unknowns, we have:

\(T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2} \\ \boxed {k=40* \frac{N}{m} }\)

If you notice any mistake in my english, please let me know, because i am not native.

Answer Link

When a spring is compressed 2.50 × 10^–2 meterfrom its equilibrium position, the total potentialenergy stored in the spring is 1.25 × 10^–2 joule.Calculate the spring constant of the spring.[Show all work, including the equation andsubstitution with units.]

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Related Questions

When a spring is compressed 2.50 × 10^–2 meter

from its equilibrium position, the total potential

energy stored in the spring is 1.25 × 10^–2 joule.

Calculate the spring constant of the spring.

[Show all work, including the equation and

substitution with units.]