A body is moving with a velocity of 72 km/h on a rough horizontal surface of coefficient of friction 0.5. If the acceleration due to gravity is 10 m/s2 find the distance covered before it comes to rest? its..urgent..
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Answered by
82
Given:
Initial velocity (u) = 72km/h = 72 x1000/3600 = 20m/s
Coefficient of friction = 0.5
g = 10m/s
Final velocity = 0
a = coeff. of friction x g
a = - 0.5 x 10
a = -5 m/s²
Now, v² - u² = 2as
0 - (20)² = 2 x -5 x s
s = - 400/-10 = 40m
Initial velocity (u) = 72km/h = 72 x1000/3600 = 20m/s
Coefficient of friction = 0.5
g = 10m/s
Final velocity = 0
a = coeff. of friction x g
a = - 0.5 x 10
a = -5 m/s²
Now, v² - u² = 2as
0 - (20)² = 2 x -5 x s
s = - 400/-10 = 40m
Answered by
10
Hlo mate
Answer:-
40 m✴✴✴✴✴✴
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