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

Compression of the spring: 0.18 m (downward)

Step-by-step 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=(m)/(\rho_w)=(5.32)/(622)=0.0086 m^3`

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

`x=(-mg+\rho Vg)/(k)=(-(5.32)(9.8)+(1000)(0.0086)(9.8))/(179)=0.18 m`

So, the spring is compressed by 0.18 m.