A stone of size 5 kg. is thrown with a velocity of 40 m/sec.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?
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Answer:
-160 N
Explanation:
m = 5 kg
u = 40 m/s
v = 0 m/s
s = 50 m
t = s/u = 50/40 s
a = (v - u)/t = (0 - 40) (4/5) = (-8) (4) = -32 m/s²
f = ma = (5) (-32) = -160 N
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