Question

A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0N

Answer 1

Answer:

12.0N

Explanation:

This problem is solved using Hooke's law which states that the extension or compression experienced by an elastic material is proportional the the force or load applied provided that the elastic limit is not exceeded.

$F=ke..................(1)$

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

$e=l_1-l_o..................(2)$

were $l_o$ is the original length of the material and $l_1$ is its total length after extension.

Given;

F = 8.0N

$l_o=3cm\\l_1=5cm\\k=?$

therefore,

$e=5-3\\e=2cm$

substituting all necessary values into equation (1), we get;

$8=k*2\\k=\frac{8}{2}\\k=4N/cm$

The force required to give the spring a total length of 6cm is thus calculated;

in this case,

$l_1=6cm\\therefore\\e=6-3\\e=3cm$

hence F is given as follows;

$F=(4N/cm) *3cm\\F=12.0N$

Answer 2

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