Hello!
using hooke's law, f s p r i n g = k 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:
\(F = K * \Delta{x}\)
\(2\:N = 4\:N/cm*\Delta{x}\)
\(4\:N/cm*\Delta{x} = 2\:N\)
\(\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}\)
simplify by 2
\(\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}\)
\(\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark\)
Answer:
d. 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)
