A counter attendant in a diner shoves a ketchup bottle with mass 0.20kg along a

Question

A counter attendant in a diner shoves a ketchup bottle with mass 0.20kg along a smooth, level lunch counter. The bottle leaves her hand with an initial velocity of 2.8m/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 1.0m before coming to rest. What are the magnitude and direction of the friction force acting on it?
part A suppose the counter attendant pushes a 0.3 kg bottle with the same initial speed on a different countertop and it travels 1.8 m before stopping what is the magnitude of the friction force from this second counter?

Explanation / Answer

using equation

v^2 = u^2 + 2as

0 = 2.8^2 + 2*a*1

a = - 3.92 m/s^2

so friction force = 3.92*0.2 = 0.784 N

direction is opposite to displacement.

part A)

v^2 = u^2 + 2as

0 = 2.8^2 + 2*a*1.8

a = - 2.18 m/s^2

friction force from this second counter = 2.18*0.3 = 0.654 N

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