What is the elongation ΔL of the spring?

Question

A light spring of force constant k = 160 N/m rests vertically on the bottom of a large beaker of water (Fig. P9.34a). A 5.00-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. What is the elongation ΔL of the spring?

Answer 1

had there been no water in the beaker, wood block would have had compression governed by
F = - k ΔL = mg = 5g
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due to water filled scenario, block will also be subjected to bouyancy force upwards - the net force on spring will decide the extension or compression.

apparent weight of block = W' = mg - [V d(water)*g]
where V = volume of block = m/d(wood)
W' = mg - m[(d(water)/d(wood)) * g]
W' = 5*9.8 - 5[(1000/650) *9.8]
W' = 49 N - 75.39N = - 26.40 N

this is upward pull on the spring>>>
restoring force = + 26.40 N downwards
so there will be extension in spring

ΔL = 26.40/k = 26.40/160 = 0.165 meter
extension
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Answer 2

Convert 5 Kg to weight in Newtons. ΔL = k X weight

Answer 3

Large Beaker