Question

If a force of 80N extends a spring of natural lenght 8m by 0.4m what will be the lenght of the spring when applied force is 100n.

Answer 1

Q1. If a force of 80N extends a spring of natural lenght 8m by 0.4m what will be the lenght of the spring when applied force is 100n.

First of all we need to find the spring constant (from Hooke's law):

$$k = \frac {F _ {1}}{x _ {1}}$$

$F_{1}$ – applied force in the 1-st case. $x_{1}$ – displacement.

So, the length of the spring in the 2-nd case will be greater its natural length:

$$x _ {2} = \frac {F _ {2}}{k} = \frac {F _ {2}}{\frac {F _ {1}}{x _ {1}}} = \frac {F _ {2} x _ {1}}{F _ {1}}$$

And the total length in 2-nd case:

$$l = l _ {0} + \frac {F _ {2} x _ {1}}{F _ {1}}$$ $$l = 8 m + \frac {1 0 0 N * 0 . 4 m}{8 0 N} = 8. 5 m$$

Answer: $l = 8.5m$