Question

A 10.-newton force compresses a spring
0.25 meter from its equilibrium position.
Calculate the spring constant of this spring.
[Show all work, including the equation and
substitution with units.]

Answer 1

     Using the Hooke's law, we have:

$F_{el}=k\Delta x \\ 10=k*0.25 \\ \boxed {k=40N/m}$

If you notice any mistake in my english, please let me know, because i am not native.

Answer 2

Answer:

Getting details from the question;

Force applied (F) = 10N

Displacement (x)= 0.25m

From Hooke's law, the force (F) needed to extend or compress a spring by some distance(displacement) x is directly proportional to that distance. That is,

F = k * x (where k is the spring constant) -------------------- (i)

Substituting the values of force and displacement got from the question into the equation (i) above, we have;

10 = k * 0.25

Making k the subject of the formula, we have;

k = 10 / 0.25

k = 40N/m

Hope this helps!