Suppose that a spring has a natural length of 25 cm, and that 8 J of work is needed to stretch from a length of 50 cm to 80 cm. How far beyond its natural length will a force of 32 N keep the spring stretched?
**I've tried calculating the spring constant using the given force and stretch distance, and using that with the 32N of force to calculate the new stretch distance, but can't seem to get it.
To solve this problem, we need to use Hooke's Law, which states that the force applied to a spring is proportional to the displacement from its natural length. The formula is given as:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement from the natural length of the spring.
Given:
Natural length (L₀) of the spring = 25 cm
Initial stretch distance (x₁) = 50 cm - 25 cm = 25 cm
Final stretch distance (x₂) = 80 cm - 25 cm = 55 cm
Work done (W) = 8 J
We can use the formula for work done on a spring to find the spring constant (k):
W = (1/2) * k * (x₂² - x₁²)
Let's substitute the given values:
8 = 0.5 * k * (55² - 25²)
Now, let's solve for k:
8 = 0.5 * k * (3025 - 625)
8 = 0.5 * k * 2400
16 = k * 2400
k = 16/2400
k = 0.0067 N/cm
Now we can use Hooke's Law to find the displacement (x) of the spring with a force of 32 N:
32 = 0.0067 * x
Let's solve for x:
x = 32 / 0.0067
x ≈ 4776.12 cm
Therefore, a force of 32 N will keep the spring stretched approximately 4776.12 cm beyond its natural length.