Question

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

Answer 1

using 3rd eq. of motion
v^2 - u^2=2aS
as v=0,u=20m/s,S=50m
o^0 - 20^2=2a50
-400=100a
a= -4m/s sq.

hence force of friction= 1kg × -4m/s sq.
=. -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✿❁═══┅✿