Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:
[tex]F = K * \Delta{x}[/tex]
[tex]2\:N = 4\:N/cm*\Delta{x}[/tex]
[tex]4\:N/cm*\Delta{x} = 2\:N[/tex]
[tex]\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}[/tex]
simplify by 2
[tex]\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}[/tex]
[tex]\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark[/tex]
Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
