When a certain rubber band is stretched a distance x, it exerts a restoring force F = ax + bx², where a and b are constants.The work done in stretching this rubber band from x = 0 to x = L is:

Question

Quizplus · Accepted Answer

The work done in stretching a spring from x = 0 to x = L is given by the formula W = (1/2)kL^2, where k is the spring constant. In this case, the restoring force is not constant, but varies with x. Therefore, we need to integrate the force over the distance x to find the work done.

The integral of the force F = ax + bx^2 from x = 0 to x = L is given by:

W = ∫[ax + bx^2] dx, from x = 0 to x = L
W = [(1/2)ax^2 + (1/3)bx^3] from x = 0 to x = L
W = (1/2)aL^2 + (1/3)bL^3

Therefore, the correct answer is E: aL^2/2 + bL^3/3.

When a Certain Rubber Band Is Stretched a Distance X