Physics: Aspring has a spring stiffness constant, k, of 440 n/m. how much must the spring be stretched to store 25 j of potential energy?

Aspring has a spring stiffness constant, k, of 440

hipstergirl225

06.01.2022

The spring is stretched by 33.7 cm.

Explanation:

It is given that,

Spring constant, k=440 N/m.

Energy stored in the stretched spring, E=25 J.

Now, Spring potential energy when it is stretched x meters is $\dfrac{1}{2}\times k \times x^2.$

Therefore, $\dfrac{1}{2}\times k \times x^2=25\\\dfrac{1}{2}\times 440\times x^2=25\\x^2=0.113\ m\\x=0.337\ m=33.7\ cm.$

The spring is stretched by 33.7 cm.

Hence, this is the required solution.

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myparentsrock17

28.09.2022

The correct answer is "1080 W".

Explanation:

Given:

Current,

I = 9.0 A

Potential difference,

V = 120 volt

As we know,

⇒ Power=IV

On substituting the values, we get

⇒ $=120\times 9.0$

⇒ $=1080 \ W$

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