A block of mass m = 0.889 kg is fastened to an unstrained horizontal spring whose spring constant is k = 89.4 N/m. The block is given a displacement of +0.128 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

To find the answers to these questions, we can use the formulas and principles of simple harmonic motion. Let's go through each question step by step:

(a) What is the force (with sign) that the spring exerts on the block just before the block is released?

The force exerted by a spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position. The equation for Hooke's Law is:

F = -kx

Where F is the force, k is the spring constant, and x is the displacement. In this case, the displacement is given as +0.128 m. Plugging in the values:

F = -(89.4 N/m)(0.128 m)
F = -11.47 N

So the force exerted by the spring just before the block is released is -11.47 N, where the negative sign indicates that the force is directed opposite to the displacement.

(b) Find the angular frequency of the resulting oscillatory motion.

The angular frequency (ω) of an oscillating system can be found using the formula:

ω = √(k/m)

Where k is the spring constant and m is the mass of the object. Plugging in the values:

ω = √(89.4 N/m / 0.889 kg)
ω ≈ 10.646 rad/s

So the angular frequency of the resulting oscillatory motion is approximately 10.646 rad/s.

(c) What is the maximum speed of the block?

The maximum speed of an object undergoing simple harmonic motion can be found using the formula:

v_max = ωA

Where v_max is the maximum speed, ω is the angular frequency, and A is the amplitude (maximum displacement) of the motion. In this case, the amplitude is given as 0.128 m. Plugging in the values:

v_max = (10.646 rad/s)(0.128 m)
v_max ≈ 1.365 m/s

So the maximum speed of the block is approximately 1.365 m/s.

(d) Determine the magnitude of the maximum acceleration of the block.

The maximum acceleration (a_max) of an object undergoing simple harmonic motion can be found using the formula:

a_max = ω^2 A

Where a_max is the maximum acceleration, ω is the angular frequency, and A is the amplitude (maximum displacement) of the motion. In this case, the amplitude is given as 0.128 m. Plugging in the values:

a_max = (10.646 rad/s)^2 (0.128 m)
a_max ≈ 18.27 m/s^2

So the magnitude of the maximum acceleration of the block is approximately 18.27 m/s^2.