## 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

1. anonymous

So 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.

2. anonymous

Note that you should integrate with respect to distance (x).

3. anonymous

thanks 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

4. anonymous

5. anonymous

oh, there is negative work?

6. anonymous

well 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