When a 2.50 - kg object is hung vertically on a certain lightspring described by Hooke’s law, the spring stretches 2.76 cm.(a) What is the force constant of the spring? (b) If the 2.50 - kgobject is removed, how far will the spring stretch if a 1.25 - kgblock is hung on it? (c) How much work must an external agentdo to stretch the same spring 8.00 cm from its unstretchedposition?

Question
Asked Dec 17, 2019
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When a 2.50 - kg object is hung vertically on a certain light
spring described by Hooke’s law, the spring stretches 2.76 cm.
(a) What is the force constant of the spring? (b) If the 2.50 - kg
object is removed, how far will the spring stretch if a 1.25 - kg
block is hung on it? (c) How much work must an external agent
do to stretch the same spring 8.00 cm from its unstretched
position?

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Expert Answer

Step 1

Given,

            Mass of the object, m=2.50kg

            Extension of the spring, x=0.0276m

Step 2

(a)

The spring weight of the body is equal to the restoring force

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kx = mg mg k=.

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Step 3

Substituting the values gives for...

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(2.50kg) (9.8m/s) k= 0.0276m =887.68N/m

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