Question

A light spring of constant 180N/m rests vertically on the bottom of a large beaker of water. A 5.71 kg block of wood of density 644 kg/m^3 is connected to to the top of the spring and the block spring system is allowed to come to static equilibrium. What is the elongation of the spring? The acceleration of gravity is 9.8m/s^2. Answer in units of cm.

Answer 1

In order to determine the elongation of the spring, proceed as follow:

Consider that the weight plus the elastic force on the block is equal to the buoyant force of the water:

W + F = E

where:

W: weight of the block = m*g = (5.71kg)(9.8m/s^2) = 55.96N

F: elastic force = kΔL

k: spring constant = 180N/m

E: buoyant force = m*g*ρw/ρb

Then, the net force on the block can be written as follow:

$\begin{gathered} F=mg(\frac{\rho_w}{\rho_b})-W \\ F=(5.71kg)(9.8\frac{m}{s^2})(\frac{1000\frac{kg}{m^3}}{644\frac{kg}{m^3}})-55.96N \\ F=86.89N-55.96N \\ F=30.93N \end{gathered}$

Next, use the previous value into the formula for the elastic force (Hooke's law) and solve for ΔL:

$\Delta L=\frac{F}{k}=\frac{30.93N}{180\frac{N}{m}}\approx0.172N$