Question

A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced (-) a
distance of 12cm and released. What is the acceleration at the instant the displacement is x = +7 cm?
a = (k / m) A or a = k x / m

Answer 1

We can find the acceleration from the Hooke's Law:

$$F=ma=-kx,$$ $$a=-\dfrac{kx}{m},$$ $$a=-\dfrac{400\ \dfrac{N}{m}\cdot 0.07\ m}{2\ kg}=-14\ \dfrac{m}{s^2}.$$

The sign minus means that when the displacement is +7 cm (downward) the acceleration is directed upward (hence, in the opposite direction to the motion of the 2-kg mass).

Answer:

$a=-14\ \dfrac{m}{s^2},$ upward.