Question

A stone of a 1 Kg is thrown in air with velocity is of a 20 m per second equation of the surface of a lake and come to rest after travelling a distance of 50 M what is a force of friction when is the stone and ice

Answer 1

hey!!!


here is your answer...


Initial velocity, u = 20 m/s
Final velocity, v = 0 ( the stone stops )
Acceleration, a = ?   ( To be calculated )
And,    Distance travelled, s = 50 m
Now,    v2 = u2 + 2as
​So,     (0)2 = (20)2 + 2 x a x 50
                0  = 400 + 100 a
          100 a = -400
                 a = -400/100
                 a = -4 m/s2
Now,        Force, F = mass x acceleration
                           F = 1 x (-4) N
                          F = -4 Newtons
Thus, the force of friction between the stone and the ice is 4 newtons. The negative sign shows that this force opposes the motion.
 
Hope you Understand !

hope it helps!

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

u = 20 m/s

v = 0 m/s

s = 50 m

According to the third equation of motion:

v^2 = u^2 + 2as

(0)^2 = (20)^2 + 2 × a × 50

a = – 4 m/s2

★The negative sign indicates that acceleration is acting against the motion of the stone.

m = 1 kg

From Newton's second law of motion:

F = Mass x Acceleration

F= ma

F= 1 × (– 4) = – 4 N

Hence, the force of friction between the stone and the ice is – 4 N.

I hope, this will help you.

Thank you______❤

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