Question

A body of mass 100g is at rest on a smooth surface .A force of 0.2N acts on it 5seconds.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 10m×g=100×10 = 1000 N

Force applied F = 0.2 N

Time (t) = 5 sec

Initial velocity (u) = 0

We now that

Force = mass \times accelerationmass×acceleration

0.2 = 0.1 \times a0.1×a

=> a = 2 m/sec^2m/sec

2

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

2

1

at

2

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

2

S = 0.5 \times 2 \times 25 = 25 mS=0.5×2×25=25m

Therefore, distance traveled by the body = 25 m

Step-by-step explanation:

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