light spring of force constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Figure a). A 4.36-kg block

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light spring of force constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Figure a). A 4.36-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 (Figure b). What is the elongation ΔL of the spring?

Physics

1 answer:

Fantom [35] · Answer 1 · 2022-08-22T23:35:22+03:00

Answer:

0.146 m

Explanation:

f = -KΔL according to Hooke's law

volume of water displaced = mass / density of block since a body will displace equal volume of its own

weight of water displaced = mass of water × acceleration due to gravity

and mass of water = volume of water / density of water

weight of water displaced = Vw × dw × g = mg (dw / dblock)

net force = mg - mg (dw / dblock) = 42.728 - 65.74 = -23.00

it will be balanced by a restoring force of 23 N

ΔL = F / k = 23 / 158 = 0.146 m