A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a

Question

3 years ago

14

t 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?

1 answer:

Kaylis [27] · Answer 1 · 2022-02-05T09:47:36+03:00

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