An elevator cab of mass m = 75 kg is descending with a speed vi = 5.0 m/s when i

Question

An elevator cab of mass m = 75 kg is descending with a speed vi = 5.0 m/s when its supporting cable begins to slip, allowing it to fall with constant acceleration a = g/5. a) During the fall through the distance d = 15 m, what is the work Wg done on the cab by gravitational force Fg ? b) During the 15 m fall, what is the work done WT done on the cab by the upward pull T of the elevator cable ? c) What is the net work done on the cab during the fall? d) What is the speed at the end of the 15 m fall?

Explanation / Answer

Force in the cart by gravitaion = mg = 750 N

Hence the work = F.d = 750 *15 =11250 J

THe force of tension in the string will be

Mg - T = ma

a is given to be g/5

T = 750*4/5 = 600 N

Hence the work done = T.d = 600*15 = 9000 J

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