Answer:
The force of friction is 622.58 N.
Explanation:
Given that,
Mass of the bottle, m = 0.27 kg
Finally it stops, v = 0
Distance traveled by the bottle, d = 1.7 mm = 0.0017 m
Let the initial speed of the bottle, u = 2.8 m/s
Let f is the force of friction is acting on the bottle. The force of friction is given by Newton's second law of motion as :
[tex]-f=ma\\\\a=\dfrac{-f}{m}.............(1)[/tex]
Using third equation of motion :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad[/tex]
[tex]v^2=\dfrac{-2fd}{m}\\\\f=\dfrac{-v^2m}{2d}\\\\f=\dfrac{-(2.8)^2\times 0.27}{2\times 0.0017 }[/tex]
f = -622.58 N
So, the force of friction is 622.58 N.
