Final answer:
When the weight is slowly lowered to its equilibrium position, it stretches a certain distance based on Hooke's law. However, when the weight is dropped from rest, the spring will stretch further before momentarily stopping and bouncing back up. The distance the spring would stretch when the weight is dropped can be found using the principle of conservation of mechanical energy.
Step-by-step explanation:
When the weight is slowly lowered to its equilibrium position, the spring will stretch a certain distance based on Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, the displacement is 3.82 cm. However, when the weight is dropped from rest, the spring will stretch further before momentarily stopping and bouncing back up. This is because the weight will experience an additional downward force due to its initial kinetic energy, which results in an increase in the total potential energy of the system.
To find the distance the spring would stretch when the weight is dropped, we can use the principle of conservation of mechanical energy. The initial potential energy of the system is equal to the final potential energy when the weight is momentarily at rest and about to bounce back up. The initial potential energy is given by the equation: PE = (1/2)kx^2, where k is the spring constant and x is the displacement. The final potential energy is zero when the weight is at its highest point and momentarily at rest. Therefore, we can set the initial potential energy equal to zero and solve for the displacement, d.
Let's use the given information to find the answer:
Given: x = 3.82 cm = 0.0382 m, k = ? (unknown)
To find k, we can use the formula: k = F/x, where F is the force exerted by the spring.
We are given the mass of the weight, but we need to calculate the force. We can use the formula: F = mg, where m is the mass and g is the acceleration due to gravity. Once we know the force, we can calculate the spring constant and then use it to find the displacement, d.
Let's calculate the spring constant first:
k = F/x = (mg)/x = (0.300 kg)(9.8 m/s^2)/(0.0382 m) = 76.1 N/m
Now, using the conservation of mechanical energy:
Initial potential energy = Final potential energy
(1/2)kx^2 = 0
(1/2)(76.1 N/m)(d)^2 = 0
d^2 = 0
d = 0 m
Therefore, when the weight is dropped, the spring will not stretch at all before momentarily stopping and bouncing back up.