physics

It takes 2.5 J of energy to stretch a helical spring by 15 cm. Determine the spring
constant.

1. 👍 0
2. 👎 0
3. 👁 76
1. PE=kx²/2
k=2•PE/ x²=
=2•2.5/0.15²=222 N/m

1. 👍 0
2. 👎 0
posted by Elena

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