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 stone and the ice.

Answer 1

[0010001111]... Hello User... [1001010101]
Here's your answer...

Mass m = 1 kg
Initial velocity u = 20 m/s
Final velocity v = 0 (because it comes to rest)
Distance S = 50 m
Acceleration a =?
By equations of motion...
v²-u² = 2aS
Since acceleration here is negative...
v²-u² = -2aS
0-20² = -2×a×50
400 = 100a
a = 4 m/s²
The stone suffers a retardation of 4 m/s²
Force of friction = ma
= 1 kg × 4 m/s²
= 4 N

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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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