Question

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density 622 kg/m3 is connected to the top of the spring and the block-spring system is allowed to come to static equilibrium.

Answer 1

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$, where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude $kx$, where $k=179N/m$ is the spring constant and $x$ is the elongation of the spring

- The buoyant force, upward, of magnitude $\rho V g$, where $\rho=1000 kg/m^3$ is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

$mg+kx-\rho V g=0$ (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

$\rho_w = 622 kg/m^3$ (wood density)

So, the volume is

$V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3$

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

$x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m$

So, the spring is compressed by 0.18 m.