a stone of 1 kg is thrown with a velocity of 20 m second across the frozen surface of the lake and cons to the rest after travelling a distance of 50 M what is the force of friction between the stone and the ice
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Answered by
4
Force = Mass x Acceleration
v² - u² = 2as
As v = 0 and s = 50m
-400 = 2*50*a
a = -4m/s²
Now put the value of a in Force formula
Force (friction) = 1*(-4) = -4N
v² - u² = 2as
As v = 0 and s = 50m
-400 = 2*50*a
a = -4m/s²
Now put the value of a in Force formula
Force (friction) = 1*(-4) = -4N
robinsingh024pbek1f:
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Answered by
3
☺ 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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