A spring with a force constant of 200 N/m is compressed by a distance of 0.4 m. A block of mass 2 kg is then placed on the compressed spring. The block is then released from rest. Calculate the maximum acceleration of the block as it is being released.
To solve this problem, we will use Hooke's law and Newton's second law of motion.
Step 1: Using Hooke's law, we can determine the force exerted by the spring when it is compressed by a distance x. Hooke's law states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
The formula for Hooke's law is given by:
F = -kx
Where F is the force exerted by the spring, k is the force constant (spring constant), and x is the displacement from the equilibrium position.
Given that the force constant of the spring is 200 N/m and the spring is compressed by a distance of 0.4 m, we can calculate the force exerted by the spring:
F = -kx
= -200 * 0.4
= -80 N
The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
Step 2: The force exerted by the spring (F) is equal to the weight of the block (mg) when the block is at equilibrium. Therefore, we can equate the two forces to find the value of the weight:
F = mg
-80 = 2 * g
Solving for g, the acceleration due to gravity:
g = -80 / 2
= -40 m/s²
Step 3: The maximum acceleration of the block occurs when the spring is fully extended. At this point, the force exerted by the spring is zero, and only the weight force acts on the block. The maximum acceleration (a) is given by Newton's second law of motion:
F = ma
-mg = ma
-mg = m * max
Solving for max, the maximum acceleration:
max = g
= -40 m/s²
Therefore, the maximum acceleration of the block as it is being released is 40 m/s² in the downward direction.