A block of mass m = 0.605 kg is fastened to an unstrained horizontal spring whose spring constant is k = 67.5 N/m. The block is given a displacement of +0.120 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest.

(a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released?


(b) Find the angular frequency ω 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.

a. What is kx?

c. Max speed of block? 1/2 m v^2=kx

solve for v

b. Isn't there a standard formula for resonant freuency with a mass on a spring?

d. max acceleration: F=ma
a=kx/m

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

F = -kx

Where F is the force, k is the spring constant, and x is the displacement.

In this case, the displacement is +0.120 m, and the spring constant is k = 67.5 N/m. Plugging these values into the equation, we get:

F = -(67.5 N/m)(0.120 m) = -8.1 N

Therefore, the force exerted by the spring just before the block is released is 8.1 N, directed in the negative x direction.

(b) The angular frequency (ω) of the resulting oscillatory motion can be determined using the formula:

ω = √(k/m)

Where k is the spring constant and m is the mass of the block.

In this case, k = 67.5 N/m and m = 0.605 kg. Plugging these values into the equation, we get:

ω = √(67.5 N/m / 0.605 kg) = √111.57 rad/s ≈ 10.574 rad/s

The angular frequency of the resulting oscillatory motion is approximately 10.574 rad/s.

(c) The maximum speed of the block can be determined using the formula:

v_max = Aω

Where A is the amplitude of oscillation (which is equal to the displacement given, +0.120 m), and ω is the angular frequency.

In this case, A = 0.120 m and ω = 10.574 rad/s. Plugging these values into the equation, we get:

v_max = (0.120 m)(10.574 rad/s) = 1.268 m/s

Therefore, the maximum speed of the block is 1.268 m/s.

(d) The magnitude of the maximum acceleration of the block can be determined using the formula:

a_max = Aω^2

Where A is the amplitude of oscillation (which is equal to the displacement given, +0.120 m), and ω is the angular frequency.

In this case, A = 0.120 m and ω = 10.574 rad/s. Plugging these values into the equation, we get:

a_max = (0.120 m)(10.574 rad/s)^2 = 14.06 m/s^2

Therefore, the magnitude of the maximum acceleration of the block is 14.06 m/s^2.

To answer these questions, we need to use the equations of motion for a mass-spring system and the principles of simple harmonic motion.

(a) The force exerted by the spring on the block can be determined using Hooke's Law, which states that the force is directly proportional to the displacement from the equilibrium position. The magnitude of the force can be calculated using the formula:

F = -kx

where F is the force, k is the spring constant, and x is the displacement. In this case, the displacement is +0.120 m, and the spring constant is 67.5 N/m.

So, the force exerted by the spring is:

F = -(67.5 N/m)(0.120 m)
= -8.10 N

The negative sign indicates that the force is in the opposite direction of the displacement, or towards the equilibrium position.

(b) The angular frequency (ω) of the resulting oscillatory motion can be found using the formula:

ω = √(k / m)

where ω is the angular frequency, k is the spring constant, and m is the mass of the block. In this case, the spring constant is 67.5 N/m, and the mass is 0.605 kg.

So, the angular frequency is:

ω = √(67.5 N/m / 0.605 kg)
= √(111.57)
≈ 10.57 rad/s

(c) The maximum speed of the block can be determined using the formula:

v_max = Aω

where v_max is the maximum speed, A is the amplitude (in this case, equal to the displacement), and ω is the angular frequency. In this case, the displacement is +0.120 m and the angular frequency is approximately 10.57 rad/s.

So, the maximum speed of the block is:

v_max = (0.120 m)(10.57 rad/s)
≈ 1.27 m/s

(d) The maximum acceleration of the block can be found using the formula:

a_max = Aω^2

where a_max is the maximum acceleration, A is the amplitude (displacement), and ω is the angular frequency. In this case, the displacement is +0.120 m and the angular frequency is approximately 10.57 rad/s.

So, the maximum acceleration of the block is:

a_max = (0.120 m)(10.57 rad/s)^2
≈ 1.60 m/s^2

So, the magnitude of the maximum acceleration of the block is approximately 1.60 m/s^2.