A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.

A 4.11 kg block of wood of density 609 kg/m3

is connected to the top of the spring and

the block-spring system is allowed to come to

static equilibrium

What is the elongation ∆L of the spring?

The acceleration of gravity is 9.8 m/s

Answer in units of cm.

Respuesta :

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

[tex]F=ke................(1)[/tex]

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

[tex]F=mg.............(2)[/tex]

m = 4.11kg

[tex]g=9.8m/s^2[/tex]

Hence,

[tex]F=4.11*9.8\\F=40.278N[/tex]

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

[tex]40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m[/tex]

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

[tex]0.2471m=0.2471*100cm=24.71cm[/tex]

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.