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A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N.How long will the spring now be, and how much work was required to stretch it that distance?

Respuesta :

isaacoranseola

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

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