An applied force varies with position according to eq. Fx = 3x3 - 5, where x is in m, and Fx in N. What is the work of this force on a body that moves from x to x = 4 m = 7 m?
To find the work done by the force on the body as it moves from x to x = 4 m, you need to calculate the definite integral of the force function Fx with respect to x over the interval [x, 4].
The work done by a force is given by the formula:
Work = ∫(F(x) dx),
where F(x) is the force function and dx represents the differential element indicating integration with respect to x.
Given that the force function is Fx = 3x^3 - 5, you can find the work done by integrating this function with respect to x from x to 4:
Work = ∫(3x^3 - 5) dx,
Now, let's find the antiderivative of the given force function:
∫(3x^3 - 5) dx = x^4 - 5x + C,
where C is the constant of integration.
Next, evaluate the antiderivative at the upper limit (4) and subtract the value at the lower limit (x):
Work = (4^4 - 5(4)) - (x^4 - 5x),
Since the lower limit is x (given in the question), the work done is:
Work = (4^4 - 5(4)) - (x^4 - 5x).
Now, substitute the values of x provided in the question, where x = 7 m:
Work = (4^4 - 5(4)) - (7^4 - 5(7)),
Calculate the expression:
Work = (256 - 20) - (2401 - 35),
Work = 236 - 2366,
Work = -2130 J.
Hence, the work done by the force on the body as it moves from x = 4 m to x = 7 m is -2130 Joules.