Question

Suppose the counter attendant pushes a 0.27 kgkg bottle with the same initial speed on a different countertop and it travels 1.7 mm before stopping. What is the magnitude of the friction force from this second counter

Answer 1

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 :

$-f=ma\\\\a=\dfrac{-f}{m}.............(1)$

Using third equation of motion :

$v^2-u^2=2ad$

$v^2=2ad$

$v^2=\dfrac{-2fd}{m}\\\\f=\dfrac{-v^2m}{2d}\\\\f=\dfrac{-(2.8)^2\times 0.27}{2\times 0.0017 }$

f = -622.58 N

So, the force of friction is 622.58 N.