Question

1. A constant force of friction of SO N is acting on a body of mass 200kg moving initially with a speed
of 15 m/s How long does the body take to stop? What distance will it cover before coming to rest?

give me pubg id​

Answer 1

Answer:

Solution➡

X = 0.3\overline{178}0.3178

10X = 3.\overline{178}3.178 .... 1 eq

10000X = 3178.\overline{178}3178.178 .... 2 eq

Now,

2nd eq - 1st eq

=> {10000}X = 3178.\overline{178}10000X=3178.178

- {10}X = 3.\overline{178}10X=3.178

=> {9990X = 3175}9990X=3175

=> X = \frac{3175}{9990}X=99903175

=> X = \frac{635}{1998}X=1998635 ✔

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