If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work is needed to stretch it 3 in. beyond its natural length?
1/2 k x^2 = 9 ft-lb
1/2 k (x/4)^2 = 9/16 ft-lb
To calculate the amount of work needed to stretch the spring 3 inches beyond its natural length, we can use the principle of conservation of energy.
The work done on the spring is equal to the potential energy stored in the spring. The potential energy stored in a stretched spring can be calculated using Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.
The potential energy (U) stored in a spring can be expressed as:
U = (1/2) * k * x^2
Where:
U is the potential energy
k is the spring constant (a measure of the stiffness of the spring)
x is the displacement from the equilibrium position
In this case, we are given the work required to stretch the spring by 1 ft, which is 9 ft-lb. From this information, we can determine the spring constant.
The work done (W) on the spring is given by:
W = U = (1/2) * k * x^2
Where x is the displacement.
Since we know the work done (W) and x, we can rearrange the equation to solve for the spring constant (k):
k = 2 * W / x^2
Using the given values, we have:
W = 9 ft-lb
x = 1 ft
Plugging in these values, we can solve for the spring constant:
k = 2 * 9 ft-lb / 1^2 ft^2
k = 18 ft-lb/ft^2
Now that we know the spring constant (k), we can calculate the work required to stretch the spring by 3 inches, which is 3/12 ft.
Using the formula for potential energy (U) and solving for the work done (W), we have:
W = U = (1/2) * k * x^2
Where:
k = 18 ft-lb/ft^2 (spring constant)
x = 3/12 ft (displacement)
Plugging in these values:
W = (1/2) * 18 ft-lb/ft^2 * (3/12 ft)^2
W = 1.125 ft-lb
Therefore, the work required to stretch the spring 3 inches beyond its natural length is 1.125 ft-lb.