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iren [92.7K]
3 years ago
7

A 550N object has a coefficient of .012 against a smooth surface. What

Physics
1 answer:
Brrunno [24]3 years ago
4 0

Vertically, the object is in equilibrium, so that the net force in this direction is

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0

where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.

Horizontally, the net force would be

∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0

where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,

<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N

so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.

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