There a five positions (A=-10cm, B=-5cm, C=0cm, D=5cm, E=5cm). A block is at rest at location C when a student pulls it to position A. The student then releases the block and it begins to oscillate. Given the block has a mass of 2kg and the spring has a constant of 40 N/m, what is the block's maximum magnitude of acceleration and at what location(s) does it occur?

D and E seem to be the same position. Please verify the wording of your question

To find the block's maximum magnitude of acceleration and its location, we can apply the principles of simple harmonic motion and Hooke's Law.

First, let's determine the equilibrium position and the amplitude of the oscillation. In this case, the equilibrium position is at C, which is 0cm. And the amplitude is the maximum displacement from the equilibrium position, which is the distance from C to either A or E. Therefore, the amplitude is 10cm.

Next, we can calculate the maximum magnitude of acceleration using the formula:

a_max = -ω^2 * A

Where ω is the angular frequency and A is the amplitude.

The angular frequency (ω) can be found by using the formula:

ω = √(k/m)

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

Plugging in the values, we have:

k = 40 N/m
m = 2 kg

So, ω = √(40 N/m / 2 kg) ≈ √20 ≈ 4.47 rad/s

Now, let's calculate the maximum magnitude of acceleration:

a_max = -ω^2 * A = -4.47^2 * 0.1 ≈ -0.2 m/s^2

Note that the negative sign indicates that the acceleration is directed towards the equilibrium position.

Lastly, let's determine the location(s) where the maximum magnitude of acceleration occurs. Since the displacement (x) and acceleration (a) in simple harmonic motion are related by the equation:

a = -ω^2 * x

We can rearrange the equation to solve for x:

x = -a / ω^2

Substituting the values, we have:

x = -(-0.2 m/s^2) / (4.47 rad/s)^2 ≈ 0.0089 m ≈ 0.89 cm

Therefore, the block's maximum magnitude of acceleration occurs at 0.89 cm (location E).