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.

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 , where m = 5.32 kg is the mass of the block and is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude , where is the spring constant and is the elongation of the spring

- The buoyant force, upward, of magnitude , where 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

(1)

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

m = 5.32 kg (mass)

(wood density)

So, the volume is

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

So, the spring is compressed by 0.18 m.

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

grav potential energy