A stone of 1 kg is thrown with a velocity of 20 metre per second across the cell surface of a lake and comes to rest after travelling a distance of what is the words of saturn between the sun and the eye
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☺ Hello mate__ ❤
◾◾here is your answer...
i think given distance is 50 m
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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