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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 (dens...Question
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?
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!!!
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!!!
Answers
Damon
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
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
julie
why is the difference in density 350? where does the 1000 come from?
Damon
water density = 1,000 kg/m^3
close enough :)
close enough :)
bobpursley
the 1000 kg/m^3 is the density of water.
Damon
I used the difference in wood and water density to get the net buoyancy.
sue
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.
Can someone explain this with the correct answer please.