(a) How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m?

Video transcript

Hi everyone in this particular practice problem. We're asked to calculate the work done by the engine where there's an engine of a suspended window cleaning lift, moving the platform of mass 125 kg. A vertical distance of six five m. The options of the work are 6.4 times 10 to the power of two Jes 8.1 times 10 to the power of three jules and 7.97 times 10 to the power of four jules. So the engine does work to actually lift the platform of vertical distance of 65 m. And to solve this problem, we will apply the energy work principle. Doing work on a system actually changes the system's energy. So that will give us a delta E system to be equals to the external work done or W external, the delta E of the system will equals to the delta K or the change in kinetic energy plus delta E thermal, which is the change in thermal energy which will equals to W E X B or to work the work external work equals to the work or the or from the gravitational uh component and engine work. And this will give our equation that is going to be very useful to solve our problem. So the delta K is the variation of the kinetic energy. And initially, so initially the platform is at rest. So initially V I equals zero m per second and then final position, the platform will also come to a rest. So V F will also be zero m per second. So because of this, then the delta K will actually be zero jule. Uh Along with this, we will also assume that there is no friction forces on the engine itself. So because assuming that there's no friction on the engine, then we can know or assuming no friction forces on the engine plus platform or essentially no friction on our overall system. Then we can know that the delta thermal energy or the change in internal energy will actually equals to also zero because of no friction. So because delta K and delta E T H both equal to zero G. So substituting dose into our equation will give us zero equals W M G plus W engine or essentially the work for or the work done by the engine will equals to the work from the gravitational forces. So because of this, we can actually start calculating um the work from the gravitational force will equals to the gravitational force dot product to the displacement or the delta Y which is essentially this displacement is only going to be in the Y component or the vertical direction and this will equals to M G uh -MG Delta y multiplied by a cosine of 180° which will actually come out to be um W engine equals M G delta Y just like. So we know all values or all necessary variables for this. So we can actually calculate this. The M is going to be 125 kg. G is 9.81 m per second squared. And the delta we uh delta Y or displacement is going to be m. So the work done by the engine will then corresponds to 7.97 times 10 to the power of four Joel. And that will be the answer to this practice problem which will correspond to option D. So option D with a work with the work done by the engine being 7.97 times 10 to the power of four, Jeel is going to be the answer. And if you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.