Answer to Question #199961 in Physics for Jes

Question #199961

If it takes 5.00 J of work to stretch a Hooke’s-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 15.0 cm.

Expert's answer

"W=0.5kx^2\\\\5=0.5(0.1)^2k\\\\k=1000\\frac{N}{m}\\\\W'=0.5kx'^2=0.5(1000)(0.15)^2=11.25\\ J"

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