Answer to Question #180775 in Physics for Miyah

Question #180775

A slinky has a spring constant of 160 N/m. How much work is done on the slinky to stretch it 0.50 meters horizontally across a table?

Expert's answer

The work is

"W=\\frac12kx^2=\\frac12\u00b7160\u00b70.5^2=20\\text{ J}."

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