Question

mass=1kg u=20m/s v=0 s=50m
use this formula--v²-u²=2as​

Answer 1

Given

The initial velocity of the stone, u= 20 m/s

The final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

Find out

The force of friction between the stone and the ice

Solution

We know the third equation of motion

v² = u² + 2as

Substituting the known values in the above equation we get,

0² = (20)² + 2(a)(50)

-400 = 100a

a = -400/100  =  -4m/s² (retardation)

We know that

F = m×a

Substituting above obtained value of a = -4 in F = m x a we get,

F = 1 × (-4) = -4N

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Answer 2

Answer:

m=1kg

u=20m/s

v=0

s=50m

v^2-u^2=2as

0^2-20^2=2×a×50

0-400=100×a

-400=100×a

-400/-100=a

a=4

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