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A vertical spring has a spring constant of

  1. newtons per meter. When an object is

attached to the bottom of the spring, the spring

changes from its unstretched length of 0.50 meter

to a length of 0.65 meter. The magnitude of the

weight of the attached object is

(1) 1.1 N (3) 50. N

(2) 15 N (4) 65 N

Verified Answer :

TaylorBayley

Hookes law states F=kX where F is the force applied, k is the spring constant, and X is the extension of the spring from its resting point.

Substituting the values in, we get:

F=100(0.65-0.5)=1000.15=15N

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