A 265-kg load is lifted 18.0 m vertically with an acceleration $a = 0.160 \textrm{ g}$ by a single cable. Determine

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Here's a free-body diagram of this load with some tension force upwards exerted by a cable, some gravity down. It's accelerating upwards; it's an acceleration of 0.160 times g so times 9.8 meters per second squared there. And it's gonna go up this height, h, 18 meters. So, the tension in the cable we can figure out from Newton's second law, and we'll say the tension force up minus the gravity down has to equal the mass times acceleration; that force is always equal to ma. And we'll bring the gravity to the right hand side so it becomes plus and gravity's mg so it's plus mg then. So, the tension force is, factor out the mass times a plus g. So that's 265 kilograms times 0.16 times 9.8 meters per second squared that gives the acceleration upwards and then plus g of 9.8 meters per second squared which is a force of 3010 newtons. And the net work done on the load is the net force multiplied by the displacement so that's mass times acceleration times displacement. 265 kilograms times 0.16 times 9.8 meters per second squared times 18 meters that it goes up and that's 7480 joules of work done on the load net. The work done by the cable on the load is the force exerted by the cable on the load multiplied by displacement. And we figured out the force in part (a); that's 3012.52 newtons, before you round it, times 18 meters which is 54200 joules and that's positive because the cable force is the tension force and it's in the same direction as the displacement. The work done by gravity is gonna be negative mg times displacement because gravity is down whereas the displacement is up so they are in opposite directions and that is why we have a negative sign there. And the force of gravity is 265 kilograms times 9.8 newtons per kilogram times 18 meters and this gives us negative 46700 joules. And if you wanted to check your work, you can make sure that this minus this makes that, which it should. And then, the final speed we'll figure out from knowing that the change in kinetic energy is equal to the net work done and the net work is mass times acceleration times displacement. We already figured out the net work done and, but you know, instead of, using a number that we have already calculated, let's just carry on and we'll plug in numbers straight from the specified by the problem. So, we have one-half mv f squared because there's no initial speed so the change in kinetic energy is equal to the final kinetic energy here. And, we have mass times acceleration times displacement. And then, we'll multiply both sides by 2, divide both sides by m and then take the square root of both sides and the final speed is square root 2 times 0.16 times 9.8 meters per second squared times 18 meters—displacement—which is 7.51 meters per second, final speed.