Question

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

Answer 1

Force = ma as per newton's second law we have m= 1kg voila..
but to find a we have to dig the 3rd eqn of motion i.e
v2=u2 + 2as

u= 20m/s and v=0 because question clearly said that its gonna stop so final velocity v=0


0=20x20 + 2.a.50

-400 = 100a

a = -4 .. negative because retardation force of friction


mark it as brailist if I you understood or else comment me for futher discussion

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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