A basket of neglible wieght hangs from a vertical spring scale of constant force 1500N/m. If you sundally put a 3.0 Kg adobe brick in the basket, find the maximum distance that the spring will stretch. If, instead, you released the brick from 1.0 m above the basket, by how much will the spring stretch at its maximum elongation?
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To find the maximum distance that the spring will stretch when a 3.0 Kg adobe brick is added, we can use Hooke's Law equation:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring from its equilibrium position
In this case, the force applied to the spring is the weight of the adobe brick, given by:
F = m * g
Where:
m is the mass of the adobe brick (3.0 Kg)
g is the acceleration due to gravity (9.8 m/s^2)
Substituting the values into the equation, we have:
F = (3.0 Kg) * (9.8 m/s^2)
F = 29.4 N
Now we can find the displacement of the spring:
29.4 N = (1500 N/m) * x
Solving for x, we get:
x = 29.4 N / 1500 N/m
x = 0.0196 meters
Therefore, the maximum distance that the spring will stretch is approximately 0.0196 meters.
Now, let's consider the scenario where the brick is released from 1.0 meter above the basket. In this case, we need to calculate the potential energy of the brick at that height. The potential energy is given by:
PE = m * g * h
Where:
m is the mass of the adobe brick (3.0 Kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height (1.0 meter)
Substituting the values into the equation, we have:
PE = (3.0 Kg) * (9.8 m/s^2) * (1.0 m)
PE = 29.4 Joules
This potential energy will be converted into elastic potential energy stored in the spring as it stretches. Therefore, we can equate the potential energy to the elastic potential energy:
PE = (1/2) * k * x^2
Where:
k is the spring constant (1500 N/m)
x is the displacement of the spring from its equilibrium position
Substituting the values into the equation, we have:
29.4 Joules = (1/2) * (1500 N/m) * x^2
Solving for x, we get:
x^2 = (2 * 29.4 Joules) / (1500 N/m)
x^2 = 0.0392 meters^2
x = √(0.0392 meters^2)
x ≈ 0.198 meters
Therefore, the spring will stretch by approximately 0.198 meters at its maximum elongation when the brick is released from 1.0 meter above the basket.