A 350 g block is attached to a vertical spring whose stiffness constant is 12 N/m . The block is released at the position where the spring is unextended.

a) What is the maximum extension of the spring?

b)How long does it take the block to reach the lowest point?

To find the maximum extension of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. The formula for Hooke's Law is F = kx, where F is the force exerted by the spring, k is the stiffness constant, and x is the extension of the spring.

a) To find the maximum extension, we need to consider the forces acting on the block. At the highest point of its motion, the block is momentarily at rest before starting its descent. Therefore, the net force on the block is zero. The forces acting on the block are the weight (mg) acting downward and the force exerted by the spring (kx) acting upward. So we have:

mg - kx = 0

Solving for x, we get:

x = mg/k

Plugging in the values, we have:

x = (0.35 kg)(9.8 m/s^2)/(12 N/m)
x = 0.285 m

Therefore, the maximum extension of the spring is 0.285 meters.

b) To find the time it takes for the block to reach the lowest point, we can use the equation of motion for simple harmonic motion. The formula for the period of oscillation (T) is given by T = 2π√(m/k), where m is the mass of the block and k is the stiffness constant of the spring.

In this case, we want to find the time it takes for the block to reach the lowest point, which is half of the period (T/2). Plugging in the values, we have:

T/2 = 2π√(m/k)/2
T/2 = π√(m/k)

Now, we can calculate the time using the given values:

T/2 = π√(0.35 kg)/(12 N/m)
T/2 = π(0.35 kg)/(12 N/m)

Using a calculator, we find that:

T/2 ≈ 0.43 seconds

Therefore, it takes approximately 0.43 seconds for the block to reach the lowest point.