A2 kg mass is attached to a vertical spring and the spring stretches 15 cm from its original relaxed position.
what is the k value of the spring?
a. 130.7 n/m
b. 9.8 n/m
c. 1.307 n/m
d. 147 n/m
e. 220.5 n/m

i know it's a but i need to know why ( in advance)

A. 130.7 N/m

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extention (d)

F = k× d ⇒ k = F/d

To find Force(F) you know it is the weight of the object. Considering g = 9.8 ms⁻²

                   k  = 2×9.8/0.15 = 130.7 Nm⁻¹

Hooke’s law,

for small deformations of an object, the displacement is directly proportional to the deforming force or load.

Under these conditions the object returns to its original shape and size upon removal of the load.

The deforming force may be applied to a solid by stretching, compressing,etc. so a spring shows an elastic behavior according to Hooke’s law

Mathematically,

applied force( F) = constant (k)* displacement in length (x)

F = kx.

The value of k depends

gets less

answer: (431,2)

step-by-step explanation: