Question

A stone of 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 the ice?

Answer 1

Answers

Initial velocity of the stone, u = 20 m/s

Final velocity of the stone, v = 0 (finally the stone comes to rest)

Distance covered by the stone, s = 50 m

According to the third equation of motion:

v2 = u2 + 2as

where, a = acceleration

(0)2 = (20)2 + 2 × a × 50

0 = 400 + 100 a

-400 = 100 a

 


Therefore, a = - 4 m/s2

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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