a construction worker at thead to haul up a 10 - m length of hanging rope. how much work will it take if the…

a construction worker at thead to haul up a 10 - m length of hanging rope. how much work will it take if the rope weighs 0.7 n/m?\nset up the integral that gives the work done\n(type an exact answer, using radicals as needed)\nthe amount of work required is\n(type an integer or a decimal.)

a construction worker at thead to haul up a 10 - m length of hanging rope. how much work will it take if the rope weighs 0.7 n/m?\nset up the integral that gives the work done\n(type an exact answer, using radicals as needed)\nthe amount of work required is\n(type an integer or a decimal.)

Answer

Explanation:

Step1: Set up the integral for work

The force (F(x)) required to lift a small segment of the rope of length (dx) located at a distance (x) from the top is (F(x)=6(10 - x)) (since the weight per unit length is (6) N/m). The work (dW) done in lifting this segment is (dW=F(x)dx = 6(10 - x)dx). The total work (W) is the integral from (x = 0) to (x=10). So, (W=\int_{0}^{10}6(10 - x)dx).

Step2: Evaluate the integral

First, expand the integrand: (6(10 - x)=60-6x). Then, integrate term - by - term. (\int(60 - 6x)dx=60x-3x^{2}+C). Using the fundamental theorem of calculus (\int_{0}^{10}(60 - 6x)dx=\left[60x-3x^{2}\right]_{0}^{10}). Substitute the upper and lower limits: ((60\times10-3\times10^{2})-(60\times0 - 3\times0^{2})). (=600 - 300-0).

Answer:

The integral is (\int_{0}^{10}6(10 - x)dx) and the amount of work required is (300) J.