A light spring of constant k = 170 N/m rests vertically on the bottom of a large beaker of water. A 5.70 kg block of wood (density = 650 kg/m3) is connected to the spring and the mass-spring system is allowed to come to static equilibrium. What is the elongation, ÄL, of the spring?

Question

im using the equation of the spring constant... F=kx F being B-weight. and i keep coming up with the wrong answer. please help me!!!

Answer 1

volume of wood = (5.7/650) m^3
difference in density = 1000-650 = 350 kg/m^3
net force up = 350 (5.7/650)(9.81) = 30.1 N

F = k x
30.1 = 170 x
x = 30.1/170 = .177 meter

Answer 2

why is the difference in density 350? where does the 1000 come from?

Answer 3

water density = 1,000 kg/m^3
close enough :)

Answer 4

the 1000 kg/m^3 is the density of water.

Answer 5

I used the difference in wood and water density to get the net buoyancy.

Answer 6

when a 14.0 kg mass hangs from a spring that has a spring constant of 550nm the spring has a length of 82 cm. determine the length of the spring before any force is applied to it

Can someone explain this with the correct answer please.