a stone of size of 1 lg is thrown with a velocity of 20ms-1 across the frozen surface of a lake and comes to rest after traveling a distance of 50m. what is the force of friction between the stone and the ice?
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Answered by
3
m= 1kg. u=20m/s. v=0. s=50. a=? f=ma 2as = v square - u square.
2×a×50= 0square -20 square
100a = -400
a= 100÷ -400
a=-0.25n
f=ma
f= 1×- 0.25
f= -0.25 N
2×a×50= 0square -20 square
100a = -400
a= 100÷ -400
a=-0.25n
f=ma
f= 1×- 0.25
f= -0.25 N
Answered by
0
☺ 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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