Question

a stone of 1 kg is thrown with a velocity of 20 m second across the frozen surface of the lake and cons to the rest after travelling a distance of 50 M what is the force of friction between the stone and the ice

Answer 1

Force = Mass x Acceleration
v² - u² = 2as
As v = 0 and s = 50m
-400 = 2*50*a
a = -4m/s²

Now put the value of a in Force formula
Force (friction) = 1*(-4) = -4N

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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