   Chapter 9, Problem 28P

Chapter
Section
Textbook Problem

A light spring of force constant k = 160 N/m rests vertically on the bottom of a large beaker of water (Fig. P9.28a). A 500-kg block of wood (density = 650 kg/m3) is connected to the spring, and the block–spring system is allowed to come to static equilibrium (Fig. P9.28b). What is the elongation ΔL, of the spring? Figure P9.28

To determine
The elongation ΔL of the spring.

Explanation
At static equilibrium, the sum of forces act on the block of wood is Fy=0BFspingmg=0 . Now, this expression is rearranged for the spring force as Fspring=Bmg=[(ρwaterVblock)m]g=[(ρwater(m/ρwood))m]g . The elongation of the spring is calculated from the definition of restoring force as shown here Fspring=kΔLΔL=Fspring/k .

Given info: Density of water is 103kg/m3 , density of wood is 650kg/m3 , mass of the block is 5.00kg , the force constant of the spring is 160N/m , and acceleration due to gravity is 9.80m/s2 .

The formula for the elongation of the spring is,

ΔL=[ρwater/ρwood1]mgk

• ρwater is density of water

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