physics home work question regarding a horizantal spring that has a spring constant of 336 NM

Question

A 2.60-kg block is attached to a horizontal spring that has a spring constant of 336 N/m. At the instant when the displacement of the spring from its unstrained length is -0.115 m, what is the acceleration a of the object? In your answer, be sure to include the proper plus or minus sign.

Answer 1

The force in the spring is F = K.X, where K is the spring constant or stiffness given as 336 N/m

and X = spring displacement = -0.115 m, Therefore F = ( 336 N/m)(-0.115 m) = -38.64 N, the force is negative and in opposite direction

From Newton second law F = m.a, where m is the mass of the object and a is the acceleration, a = F/m

Now F = -38.64 N or -38.64 kg.m/s² , then a = (-38.64 kg.m/s²)/(2.6 kg) = -14.86 m/s², note that the acceleration is also negative and in the same direction as the force.