A stone of mass 0.5 kg is thrown with a velocity of 10 m/s across the frozen surface of a lake and it comes to rest after travelling a distance of 100 m. What is the magnitude of force of friction between the stone and ice?
Choose one:
3 N
4 N
0.25 N
0.75 N
i will mark the brainliest whoever answers it fist
Answers
We will use the following equation of motion :
V² = U² - 2as
Where V is the final speed and u the initial velocity and S is the displacement
In this case our U is 10m/s and V is 0. The S is 100.
0 = 100 - 2 × 100 × a
0 = 100 - 200a
100 = 200a
a = 0.5m/s²
Force = mass × Acceleration
Force = 0.5 × 0.5 = 0.25 N
☺ Hello mate__ ❤
◾◾here is your answer...
u = 10 m/s
v = 0 m/s
s = 100 m
According to the third equation of motion:
v^2 = u^2 + 2as
(0)^2 = (10)^2 + 2 × a × 100
a = – 0.5 m/s2
★The negative sign indicates that acceleration is acting against the motion of the stone.
m = 0.5 kg
From Newton's second law of motion:
F = Mass x Acceleration
F= ma
F= 0.5 × (– 0.5) = – 0.25N
Hence, the force of friction between the stone and the ice is – 0.25N.
I hope, this will help you.
Thank you______❤
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