A variable force of
7x−2
pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from
x = 1
ft to
x = 19
ft. (Round your answer to two decimal places.)
7x^-2*
To calculate the work done in moving the object from x = 1 ft to x = 19 ft, we need to find the integral of the variable force function over the given range.
The work done is defined as the integral of force with respect to displacement:
Work = ∫(F(x) dx)
Given that the variable force function is 7x - 2 pounds, we can substitute this into the integral:
Work = ∫((7x - 2) dx)
To find the work, we need to evaluate this integral over the interval [1, 19]:
Work = ∫[1,19] (7x - 2) dx
Integrating the function, we get:
Work = [ (7/2)x^2 - 2x ] evaluated from 1 to 19
Substituting the limits of integration:
Work = [(7/2)(19)^2 - 2(19)] - [(7/2)(1)^2 - 2(1)]
Work = [(7/2)(361) - 38] - [(7/2)(1) - 2]
Work = [1267/2 - 38] - [7/2 - 2]
Work = 1267/2 - 38 - 7/2 + 2
Simplifying, we get:
Work = 1267/2 - 7/2 - 38 + 2
Work = 1260/2 - 5/2 - 36
Work = 630 - 5/2 - 36
Work = (630 * 2 - 5)/2 - 36
Work = (1260 - 5)/2 - 36
Work = 1255/2 - 36
Work = 1255/2 - 72/2
Work = (1255 - 72)/2
Work = 1183/2
Finally, rounding the answer to two decimal places:
Work ≈ 591.50 pounds-feet
Therefore, the work done in moving the object from x = 1 ft to x = 19 ft is approximately 591.50 pounds-feet.
To calculate the work done in moving the object, we need to integrate the force function over the given interval.
The work done is given by the formula:
W = ∫F(x)dx
First, let's find the antiderivative of the force function F(x) = 7x - 2:
∫(7x - 2)dx = (7/2)x^2 - 2x + C
Now, we can find the work done by evaluating the integral from x = 1 to x = 19:
W = ∫[1,19](7/2)x^2 - 2x + C dx
W = [(7/2)(1)^2 - 2(1) + C] - [(7/2)(19)^2 - 2(19) + C]
W = [(7/2) - 2 + C] - [(7/2)(361) - 38 + C]
W = (7/2 - 2) - (7/2)(361) + 38
W = (7/2 - 2) - (7/2)(361) + 38
W = -6965.5 lb-ft (rounded to two decimal places)
Therefore, the work done in moving the object from x = 1 ft to x = 19 ft is approximately -6965.5 lb-ft.
integral of F dx
assuming that F is in exactly the direction of motion
7 integral dx/x^2 at 19 - at 1
-7 (1/x) at 19 - at 1
7/1 - 7/19 = 12/19