Question

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.
a. 4 cm
b. 8 cm
c. 2 cm
d. 1/2 cm.

Answer 1

Answer: The correct answer is Option d.

Explanation:

Formula for Hooke's Law states that:

$F_{spring}=kx$

where, F = Force applied = 2N

k = spring constant = 4 N/cm

x = displacement of the spring from equilibrium position = ? cm

Putting values in above equation, we calculate the displacement or distance of the spring, we get:

$2N=4N/cm\times x\\\\x=\frac{1}{2}cm$

Hence, the correct option is Option d.

Answer 2

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! =)