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anonymous
 5 years ago
a heavy rope 60 ft long weighs .7 lb/ft and hangs over the edge of a building 150 ft high how much work W is done
anonymous
 5 years ago
a heavy rope 60 ft long weighs .7 lb/ft and hangs over the edge of a building 150 ft high how much work W is done

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So basically we know that \[\int\limits_{}^{} F = W\] F=MA M= mass, A=acceleration The force being the total weight of the rope for any distance that it has been pulled up. To model an equation that represents the weight of the rope for any distance that it has been pulled up (x) we will start with the Total wight of the rope, minus the weight of the rope that has been pulled up. The first part of our Force equation should look like this: .7(60) That is the total wight of the rope. Now we must subtract the rope that has been pulled up already for a distance x.: .7(60).7(60)(x/60) Cancelling the 60's in the second part you get: F = .7(60).7x Now that you have your force equation, you integrate over the distance. The MOST distance that ANY point on the rope will have to travel is theoretically 60 ft. Your limits of integration are 0 to 60.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note that you should integrate with respect to distance (x).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for your replies. so dw is a sliver of work. we are given that a chunk of rope is .7 lb/ft and dx is our foot. so we have dW = force at x * distance covered = [.7 dx ] * x , since x will range from 0 to 60 thinking of the rope as upside down

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. Your question doesn't ask for what kind of work being done though; this would impact the sign of your answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, there is negative work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well since the force is positive throughout, we might as well call it positive work. here are the questions by the way http://i51.tinypic.com/295zq7r.jpg
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