A $$1000$$ kg elevator rises from rest in the basement to the fourth floor, a distance of $$20$$ m. As it passes the fourth floor its speed is $$4$$ $$m/$$$$sec$$. There is a constant frictional force of $$600$$ N. Then percentage of total work done is lost due to friction nearly is

Question

A $$1000$$ kg elevator rises from rest in the basement to the fourth floor, a distance of $$20$$ m. As it passes the fourth floor its speed is $$4$$ $$m/$$$$sec$$. There is a constant frictional force of $$600$$ N. Then   percentage  of total work done is lost due to friction nearly is

Infinity Learn · Accepted Answer

total work done $$=$$ work done against $$gravity+work$$ done against friction $$+$$ increase in kinetic energy                         $$=$$$$mgh$$$$+Ffriction$$×h $$+12mv2$$ $$=1000$$×$$9.8$$×$$20+600$$×$$20$$ $$+121000$$×$$16$$     $$=1039.8$$×$$20$$ $$+$$ $$12$$ $$+8$$  $$=103$$×$$216$$     % of work done against friction   $$12216$$×$$100=5.5$$

A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4 m/sec. There is a constant frictional force of 600 N. Then percentage of total work done is lost due to friction nearly is

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