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Integrating Force to find Work

Integrating Force to find Work. Work, measured in either joules(Newton-meters) or foot-pounds, is the product of the force necessary to move and object and the displacement or distance of the object caused by the force. It is given by the equation W = F  d.

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Integrating Force to find Work

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  1. Integrating Force to find Work

  2. Work, measured in either joules(Newton-meters) or foot-pounds, is the product of the force necessary to move and object and the displacement or distance of the object caused by the force. It is given by the equation W = Fd If a 10 lb block is pushed 4 feet across a floor, the work done in pushing the block would be 40 foot-lbs. If a 20 kg block is pushed 3 meters across a floor, the work done in pushing the block would be W = Fd = (20 kg 9.8 m/s2)(3 meters) = 588 Newton-meters. The difference here is that with metric weight measurements, we have to multiply by 9.8 m/s2

  3. Suppose that the force required to move an object x meters across the floor decreases as the object gains speed so that the force can be given by the function F(x) =Ce0.43x. If the force required to budge the object from rest is 49 N, find C. At any point x, the force required is 49e0.43x and the over a small distance x the work done would be W = 49e0.43xx. Over a distance from 0 to 6 meters we can approximate it by adding all the small x partitions so that we get As k goes to infinity, this summation becomes  105.32 joules (or Newton-meters)

  4. …in which F(x) = a force function …but there are cases in which dx can be part of the force or weight of an object being moved…

  5. Promise or we’ll leave you down there… You know what we want… Muhahahahahahaaaa Tell us what we want to hear…

  6. Tina and Danielle Yee have Mr. Murphy (weighing 82 kg) dangling from a rope, 25 m long and weighing 5 kg/m , off the edge of a building. When Mr. Murphy finally promises a lifetime supply of cookies on demand, they hoist him back up to the top of the building. How much total work is done by Tina and Danielle raising both the rope and Mr. Murphy to the top of the building? The Mr. Murphy part is easy… W = F  d = weight  distance W = F  d = 82 kg (9.8 m/s2) 25 m W = 20,090 j

  7. Ellie and Lisa have Mr. Murphy (weighing 82 kg) dangling from a rope, 25 m long and weighing 5 kg/m, off the edge of a building. When Mr. Murphy finally promises to bake them a lifetime supply of cookies, they hoist him back up to the top of the building. How much total work is done by Ellie and Lisa raising both the rope and Mr. Murphy to the top of the building? The rope part is a little different… x dx A small piece of the rope with length dx cm …and weighing 5 kg/m  dx m …is pulled up by a distance of x meters

  8. Tina and Danielle (Yee, of course) have Mr. Murphy (weighing 82 kg) dangling from a rope, 25 m long and weighing 5 kg/m, off the edge of a building. When Mr. Murphy finally promises to bake them a lifetime supply of cookies, they hoist him back up to the top of the building. How much total work is done by Tina and Danielle raising both the rope and Mr. Murphy to the top of the building? The rope part is a little different… Work (for one small piece of rope) = Weight  Distance Don’t forget about gravity 5 kg/m  dx m x meters =15,312.5 j The sum of all the little pieces of rope Distance Force (or in this case the weight) of one little piece of the rope

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