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A 2.0- kg box is moving at constant speed of 2.0 m/s on a frictionless horizontal...

A 2.0- kg box is moving at constant speed of 2.0 m/s on a frictionless horizontal floor. You apply a horizontal force with a magnitude of 4.0 N to make it stop. How long should you apply this force?

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Answer #1

F = 4 N

v= 2 m/s

m= 2kg

newton's second law of motion, impulse = change in momentum

F x t = m dv

= m ( v2 - v1 )

4 x t = 2 (2-0)

t = 1 s

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