   Chapter 5, Problem 20P

Chapter
Section
Textbook Problem

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?

(a)

To determine
The force constant of the spring.

Explanation

Given Info:

The mass of the object is 2.50kg .

The length that the spring stretched is 2.76m .

Formula to calculate the force constant of the spring is,

k=Fgd

• Fg is the force excreted on the spring due to gravity
• d is the length that the spring is stretched

Since, the spring is stretched only by the weight of the suspended object,

Equation (I) gives as,

k=mgd

• m is mass of the object
• g is acceleration due to gravity

Substitute 2.50kg for m, 9.8m/s2 for g and 2

(b)

To determine
The spring stretch, if the 2.50kg mass is replaced with 1.25kg mass.

(c)

To determine
The work done required to stretching the spring to 8.0cm .

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