Question

A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice

Answer 1

Mass of stone m = 1kg

U = 20ms-1

V = 0

Distance = 50m

V^2 – U^2 = 2as

0^2 – 20^2 = 2a * 50

A = -400/100 =  -4ms^-2

Force of friction , F = ma = 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______❤

✿┅═══❁✿ Be Brainly✿❁═══┅✿