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?

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JemdetNasr JemdetNasr · Answer 1 · 2019-12-21T12:38:06+01:00

Answer:

1.176 N

Explanation:

[tex]m[/tex] = mass of the bottle = 0.30 kg

[tex]v_{o}[/tex] = initial speed of the bottle = 2.8 m/s

[tex]v[/tex] = final speed of the bottle = 0 m/s

[tex]d[/tex] = stopping distance traveled = 1.0 m

[tex]f[/tex] = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

[tex]- 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[/tex] N

direction :

frictional force acts in opposite direction of motion.