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 and the ice?

Class 9 Cbse​

Answer 1

$\huge\bf\red{\mid{\overline {\underline {Answer}}} \mid}$

Given,

Mass, m = 1 kg

Distance, s = 50 m

Initial velocity, u = 20 m/s

Final velocity, v = 0 m/s

Acceleration, a =?

Force, F =?

Now, 2as = v^2 - u^2

=> 2 × a × 50 = 0^2 - (20)^2

=> 100 a = - 400

=> a = - 400/100

=> a = - 4

∴ F = ma = 1 × (-4) = - 4N

Hence the force of friction between the stone and the ice will be :

$\boxed {\Huge\bold{-4N}}$

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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