Question

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.

A.) 2 cm
B.) 1/2 cm
C.) 4 cm
D.) 8 cm

Answer 1

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:

`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:

B.) 1/2 cm

_______________________

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