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3 years ago

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the

Physics

2 answers:

makkiz [27]3 years ago

8 0

Answer:

31.25

Explanation:

The formula for the potential energy of a spring is $\frac{1}{2}kx^2$ , where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get $\frac{1}{2}\cdot 250\cdot 0.25=31.25$ . Hope this helps!

hodyreva [135]3 years ago

8 0

31.25 hope this helps:)

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3 years ago

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