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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..