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?

Question

Physics

Thornton

30 March, 23:37

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?

+3

Answers (1)

Know the Answer?

Skylar Sampson · Answer 1

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