A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answers
Answered by
55
M = 1kg
u = 20m/s. v = 0m/s.
s(distance travelled) = 50m
using third equation of motion
v²=u²+2as
0² = (20)²+2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
F = m×a
F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)
hope this helps
u = 20m/s. v = 0m/s.
s(distance travelled) = 50m
using third equation of motion
v²=u²+2as
0² = (20)²+2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
F = m×a
F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)
hope this helps
Answered by
11
☺ 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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