A force of 4 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?
This should clear it up for you...
http://math.stackexchange.com/questions/543655/using-hookes-law-to-find-the-work-required-to-stretch-a-spring
F = k x
4 = k (.1)
k = 40 pounds/foot
Energy = Work = (1/2) k x^2
= (1/2)(40)(.6)^2 foot pounds
to get that with calculus
work in = F dx
= k x dx
= k x^2/2 + constant
if W = 0 when x = 0 than the constant is 0
so
W = (1/2) k x^2
To find the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length, we need to calculate the change in potential energy of the spring.
The potential energy of a spring is given by the formula:
Potential Energy = (1/2) * k * x^2
Where:
- k is the spring constant, which measures the stiffness of the spring
- x is the displacement of the spring from its equilibrium position
First, we need to find the spring constant, k. The spring constant is determined by the relationship between force and displacement.
We know that a force of 4 pounds is required to hold the spring stretched 0.1 feet beyond its natural length. So, we can set up the following equation to find k:
4 pounds = k * 0.1 feet
Now we can solve for k:
k = 4 pounds / 0.1 feet = 40 pounds/feet
Now that we have the spring constant, we can find the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length.
Let's denote the initial displacement (natural length) of the spring as x1 = 0 feet, and the final displacement (0.6 feet beyond the natural length) as x2 = 0.6 feet.
The work done in stretching the spring from x1 to x2 is given by the difference in potential energies:
Work = Potential Energy at x2 - Potential Energy at x1
Potential Energy at x2 = (1/2) * k * x2^2
Potential Energy at x1 = (1/2) * k * x1^2
Substituting the values:
Potential Energy at x2 = (1/2) * 40 pounds/feet * (0.6 feet)^2
Potential Energy at x1 = (1/2) * 40 pounds/feet * (0 feet)^2 = 0
Work = (1/2) * 40 pounds/feet * (0.6 feet)^2 - 0
Calculating the expression:
Work = (1/2) * 40 pounds/feet * (0.6 feet)^2
= (1/2) * 40 pounds/feet * 0.36 feet^2
= 0.5 * 40 pounds/feet * 0.36 square feet
= 7.2 foot-pounds
Therefore, the work done in stretching the spring from its natural length to 0.6 feet beyond its natural length is 7.2 foot-pounds.