Question

A body mass of 100g is at rest On a smooth surface. A force of 0.2N acts on it for 5 sec. Calculate the distance travelled by the body.

Answer 1

Answer: The distance travelled

25m

Given that mass = 100g = 0.1kg

       Weight = $m \times g = 100 \times 10$= 1000 N

       Force applied F = 0.2 N

               Time (t) = 5 sec

       Initial velocity (u) = 0

       We now that

               Fore = $mass \times acceleration$

               0.2 = $0.1 \times a$

              => a = 2 $m/sec^2$

$S\quad =\quad ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

               $S = 0 + 0.5 \times 2 \times (5)^2$

               $S = 0.5 \times 2 \times 25 = 25 m$

Therefore, distance traveled by the body = 25 m

Answer 2

Answer:

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