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

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} = (2\:\diagup\!\!\!\!\!N)/(4\:\diagup\!\!\!\!\!N/cm)`

simplify by 2

`\Delta{x} = (2)/(4)(/2)/(/2)`

`\boxed{\boxed{\Delta{x} = (1)/(2)\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark`

Answer:

d. 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

Answer 2

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* x\\\\x=(1)/(2)cm`

Hence, the correct option is Option d.