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?
To find the maximum distance that the spring will stretch when the adobe brick is placed in the basket, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed.
The formula for Hooke's Law is:
F = k * x
Where:
F is the force applied to the spring (in newtons),
k is the spring constant (in newtons per meter),
and x is the displacement or stretch of the spring (in meters).
In this case, the force applied to the spring is the weight of the adobe brick, which is given by:
Force = mass * gravity
Since the weight of the adobe brick is acting in the downward direction, we need to use the negative value for force. Therefore:
Force = - mass * gravity
Substituting the values given:
mass = 3.0 kg
gravity = 9.8 m/s^2
Force = - (3.0 kg * 9.8 m/s^2) = -29.4 N
The spring constant is given as 1500 N/m.
Now, we can rearrange the Hooke's Law formula to solve for the displacement (x):
x = F / k
Substituting the values:
x = (-29.4 N) / (1500 N/m)
x ≈ -0.0196 m
Therefore, the spring will stretch by approximately 0.0196 meters (or 1.96 centimeters) when the adobe brick is placed in the basket.
If the adobe brick is released from 1.0 m above the basket, we can use the principle of conservation of energy to determine the maximum stretch of the spring.
The potential energy of the brick at a height h is given by:
Potential Energy = mass * gravity * height
Since the brick is released from rest, its initial potential energy is fully converted into elastic potential energy stored in the spring.
Therefore:
Potential Energy of the spring = (1/2) * k * x^2
Equating these two expressions for potential energy:
mass * gravity * height = (1/2) * k * x^2
Substituting the values:
mass = 3.0 kg
gravity = 9.8 m/s^2
height = 1.0 m
k = 1500 N/m
3.0 kg * 9.8 m/s^2 * 1.0 m = (1/2) * 1500 N/m * x^2
29.4 N * 1.0 m = 750 N/m * x^2
x^2 = (29.4 N * 1.0 m) / (750 N/m)
x^2 ≈ 0.0392 m^2
Taking the square root of both sides:
x ≈ 0.198 m
Therefore, the spring will stretch by approximately 0.198 meters (or 19.8 centimeters) at its maximum elongation when the brick is released from a height of 1.0 meter above the basket.