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A vertical spring has a spring constant of
100. 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

Respuesta :

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)=100*0.15=15N

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