A force of 10 pounds stretches a spring 2 inches. Find the work done in stretching this spring 3 inches beyond its natural length
Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is:
F = -kx (1)
where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get:
-10 = -k(2)
or
k = 5 lb/in
work is the 1st integral of force with respect to x, which would be:
W = (1/2)kx^2
= (1/2)5(3)^2
= 22.5 in-lb
To find the work done in stretching the spring 3 inches beyond its natural length, we can use the formula for work:
Work = Force * Displacement * cos(theta)
In this case, the force applied is 10 pounds and the displacement is 3 inches. However, we need to find the angle (theta) between the force and the displacement in order to calculate the work.
Since the force and displacement are in the same direction (acting along the length of the spring), the angle between them is 0 degrees. The cosine of 0 degrees is 1, so we can simplify the formula to:
Work = Force * Displacement
Substituting the given values:
Work = 10 pounds * 3 inches
To perform the calculation, we need to convert pounds to Newtons and inches to meters, since the SI units are typically used for calculations:
1 pound ≈ 4.44822 Newtons
1 inch ≈ 0.0254 meters
So,
Work = 10 pounds * 4.44822 Newtons/pound * 3 inches * 0.0254 meters/inch
Simplifying this expression gives us:
Work ≈ 13.34466 Newton-meters
Therefore, the work done in stretching the spring 3 inches beyond its natural length is approximately 13.34466 Newton-meters.