A spring with a force constant 0f 2.50 kN/M is set upright, and a wood platform of negligible
weight is fastened to its upper end. A 2.50 kg mass is gently placed on the platform and released.
How far does the spring compress? What is the final equilibrium position?
To determine how far the spring compresses and the final equilibrium position, we need to apply Hooke's Law and the concept of equilibrium.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be represented as:
F = -kx
Where F is the force exerted by the spring, k is the force constant (also known as the spring constant), and x is the displacement from the equilibrium position.
Given:
Force constant (k) = 2.50 kN/m = 2.50 * 1000 N/m = 2500 N/m
Mass (m) = 2.50 kg
To find the amount of compression (displacement), we can use the following equation:
F = -kx
Since the weight of the mass acts downwards, it will be balanced by the upward force exerted by the spring once it reaches equilibrium. At this point, the net force acting on the mass becomes zero. Therefore:
Weight (W) = mg
Force exerted by the spring (F) = kx
mg = kx
Rearranging the equation:
x = mg / k
x = (2.50 kg) * (9.81 m/s²) / (2500 N/m)
Calculating:
x = 0.0981 m
So, the spring compresses by approximately 0.0981 meters or 9.81 centimeters.
To determine the final equilibrium position, we need to consider that the weight of the mass is balanced by the force exerted by the spring. Since the weight acts downwards, the upward spring force will be equal to the weight of the mass.
Weight (W) = mg
Force exerted by the spring (F) = kx
mg = kx
Solving for x:
x = mg / k
x = (2.50 kg) * (9.81 m/s²) / (2500 N/m)
Calculating:
x = 0.0981 m
Therefore, the final equilibrium position is at a displacement of 0.0981 meters above the unfixed position of the spring.