A force of 1 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 1.1 feet beyond its natural length

Given Information:

Force = f = 1 pound

Stretched length = x = 0.1 ft

Required Information:

Work done = W = ?

Work done = 6.05 ft.lb

Explanation:

From the Hook's law we know that

f(x) = kx

Where f is the applied force, k is spring constant and x is length of spring being stretched.

k = F/x

k = 1/0.1

k = 10 lb/ft

f(x) = 10x

The work done is given by

W = ∫ f(x) dx

Where f(x) = 10x and limits of integration are (1.1, 0)

W = ∫ 10x dx

W = 10*x²/2

W = 5x²

Evaluating the limits,

W = 5(1.1)² - 5(0)²

W = 6.05 - 0

W = 6.05 ft.lb

Therefore, 6.05 ft.lb work has been done in stretching the spring from its natural length to 1.1 feet beyond it's natural length.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

good for her? ? where is the question?