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Q.

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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a

4

b

5.5

c

6.2

d

12

answer is B.

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Detailed Solution

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total work done = work done against gravity+work done against friction + increase in kinetic energy

$= m g h + F_{f r i c t i o n} \times h + \frac{1}{2} m v^{2} = 1000 \times 9.8 \times 20 + 600 \times 20 + \frac{1}{2} 1000 \times 16 = 10^{3} (9.8 \times 20 + 12 + 8) = 10^{3} \times 216 % o f w o r k d o n e a g a i n s t f r i c t i o n \frac{12}{216} \times 100 = 5.5$

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