Question

A block of mass m = 0.59 kg is attached to a spring with force constant 128 N/m is free to move on a frictionless, horizontal surface as in the figure below. The block is released from rest after the spring is stretched a distance A = 0.13 m. (Indicate the direction with the sign of your answer. Assume that the positive direction is to the right.)

    

    (a) At that instant, find the force on the block.
          N
    
    (b) At that instant, find its acceleration.
          m/s²

Answer 1

m = mass of the block = 0.59 kg
k = spring constant = 128 N/m
A = stretching of the spring = 0.13 m

The force by the spring on the block is found by Hooke's law:
F = - k A
F = - (128 N/m) (0.13 m)
F = -16.64 N
F = -17 N (to 2 sf)

The acceleration of the block is found by Newtons second law:
F = m a
a = F / m
a = (-16.64 N) / (0.59 kg)
a = -28.20 m/s