A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and spring B has a spring constant of 5 N/m.

By Hooke’s law, the stretch is the force divided by the spring constant. Spring A stretches 60 ÷ 4 = 15 m, while spring B stretches 60 ÷ 5 = 12 m. Since both springs start at the same length, spring A ends up 15 − 12 = 3 m longer, matching choice D.

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