Question

A stone of 1 kg is thrown with a velocity of 20 m/s 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 and the ice.

Answer 1

first we have to calculate retardation from the equation v^2=u^2+2as where v is 0
u get retardation=4
so force of friction is 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✿❁═══┅✿