A certain spring stretches 4.6 cm when it
supports a mass of 0.55 kg .
If the elastic limit is not reached, how far
will it stretch when it supports amass of 8 kg ?
Answer in units of cm
d = (m2/m1) * d1,
d = (8kg/0.55kg) * 4.6cm = 75.3cm.
To solve this problem, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it. Mathematically, Hooke's Law can be represented as:
F = k * x
Where:
F is the force applied to the spring (measured in Newtons)
k is the spring constant (measured in Newtons per meter)
x is the extension of the spring (measured in meters)
In this case, we are given the extension of the spring and the mass it supports. We can use this information to find the spring constant and then use it to calculate the extension for a different mass.
Given:
Extension (x1) = 4.6 cm = 0.046 m
Mass (m1) = 0.55 kg
We know that the force applied to the spring (F1) can be calculated using the formula: F1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
F1 = 0.55 kg * 9.8 m/s^2
F1 = 5.39 N
Now, let's calculate the spring constant (k) using Hooke's Law.
F1 = k * x1
5.39 N = k * 0.046 m
Solving for k, we can write:
k = 5.39 N / 0.046 m
k = 117.17 N/m
Now we can use the spring constant (k) to find the extension of the spring for a different mass.
Mass (m2) = 8 kg
Using the formula F2 = m2 * g, we can calculate the force applied to the spring.
F2 = 8 kg * 9.8 m/s^2
F2 = 78.4 N
Now, use Hooke's Law to calculate the extension (x2).
F2 = k * x2
78.4 N = 117.17 N/m * x2
Solving for x2, we can write:
x2 = 78.4 N / 117.17 N/m
x2 ≈ 0.67 m
Finally, we can convert the extension from meters to centimeters:
x2 = 0.67 m * 100 cm/m
x2 ≈ 67 cm
Therefore, the spring will stretch approximately 67 cm when supporting a mass of 8 kg.