A problem by OJAS JAIN
An elevator cab of mass m=500kg is descending with speed v=4m/s when its supporting cable begins to slip, allowing it to fall with constant acceleration a=g/5.During the 12 m fall,let the work done on the cab by the upward pull of the elevator cable be x,the net work W done on the cab during the fall be y.What is the value of |x|+y in kilojoules to the nearest integer?
The answer is 59.
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1 solution
The work done by the upward pull of elevator cable=m(g-a)h =500 4g 12/5 =-4.7 10^4 joules Now,the network done will be=work done by upward pull done by elevator cable+work done by gravitational force =4.7 10^4+mgdcos0 =-4.7 10^4+500 9.8 12 1 =5.88 10^4-4.7 10^4 =1.18*10^4 joules Therefore,|x|=47 kilojoules and y=11.8 kilojoules Therefore,|x|+y=58.8 kilojoules Therefore,answer is 59 kilojoules