Question

A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity 2.8 m/s. As it slides, it slows down because of the horizontal friction force exerted on it by the countertop. The bottle slides a distance of 1.0 m before
coming to rest. What are the magnitude and direction of the friction force acting on it?

Answer 1

Answer:

1.176 N

Explanation:

$m$ = mass of the bottle = 0.30 kg

$v_{o}$ = initial speed of the bottle = 2.8 m/s

$v$ = final speed of the bottle = 0 m/s

$d$ = stopping distance traveled = 1.0 m

$f$ = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

$- f d = (0.5) m (v^{2} - v_{o}^{2} )\\- f (1) = (0.5) (0.30) (0^{2} - 2.8^{2} )\\-f = - 1.176 \\f = 1.176$ N

direction :

frictional force acts in opposite direction of motion.