A 5.0-kg object is supported by an aluminum wire of length 2.1m and diameter 1.7mm .
Part A
How much will the wire stretch?
To determine how much the aluminum wire will stretch under the weight of the 5.0 kg object, we can use Hooke's Law. Hooke's Law states that the amount of stretch or deformation of an elastic material is directly proportional to the force applied to it.
The formula for Hooke's Law is:
F = k * ΔL
Where:
F is the force applied to the material,
k is the spring constant or stiffness of the material, and
ΔL is the change in length or stretch of the material.
First, let's calculate the force exerted by the 5.0 kg object due to gravity.
The formula for gravitational force is:
F = m * g
Where:
m is the mass of the object,
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we have:
F = 5.0 kg * 9.8 m/s^2
F = 49 N
Now, we need to determine the spring constant or stiffness of the aluminum wire. This can be done by using the formula:
k = (π * (d^2)) / (4 * L * Y)
Where:
d is the diameter of the wire,
L is the length of the wire,
Y is the Young's modulus of the material (for aluminum, Y = 7.0 x 10^10 N/m^2).
Plugging in the values, we have:
d = 1.7 mm = 0.0017 m
L = 2.1 m
Y = 7.0 x 10^10 N/m^2
Now, let's calculate the spring constant:
k = (π * (0.0017 m^2)) / (4 * 2.1 m * 7.0 x 10^10 N/m^2)
Once we have the spring constant, we can use it to find the change in length or stretch of the wire:
ΔL = F / k
Plugging in the values, we have:
ΔL = 49 N / k
By calculating the spring constant and dividing the force applied by the spring constant, we can find the amount the wire will stretch.