If the work required to stretch a spring 1 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in. beyond its natural length?
To find the work needed to stretch the spring by 6 inches, we can use the concept of spring potential energy.
The potential energy stored in a spring is given by the equation:
PE = (1/2) * k * x^2
Where PE is the potential energy, k is the spring constant, and x is the displacement from the natural length of the spring.
In this case, we are given that the work required to stretch the spring by 1 ft (12 inches) is 6 ft-lb. To find the spring constant (k), we use the formula:
Work = PE = (1/2) * k * x^2
Since the work is given in foot-pounds and the displacement (x) is in feet, we can directly substitute the given values into the equation:
6 ft-lb = (1/2) * k * (12 ft)^2
Simplifying the equation:
6 ft-lb = 6 * k * 144 ft^2
Dividing both sides of the equation by 6 * 144 ft^2:
1 ft-lb = k * 24 ft
Therefore, the spring constant (k) is 1/24 ft-lb/ft, or 1/24 lb/ft.
Now we can calculate the work needed to stretch the spring by 6 inches (0.5 feet). Using the potential energy equation:
PE = (1/2) * k * x^2
Substituting the values:
PE = (1/2) * (1/24 lb/ft) * (0.5 ft)^2
Simplifying the equation:
PE = (1/2) * (1/24 lb/ft) * (0.25 ft^2)
PE = 0.005 lb-ft
Therefore, the work needed to stretch the spring by 6 inches beyond its natural length is 0.005 foot-pounds.