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ludmilkaskok [199]
3 years ago
14

A 3.0 kg box is at rest on a table. The static friction coefficient us between the box and table is 0.40, and

Physics
1 answer:
Oduvanchick [21]3 years ago
7 0

While the box is at rest, the net vertical force acting on it is

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where <em>n</em> = magnitude of the normal force (table pushing up on the box) and <em>w</em> = weight of the box. So we have

<em>n</em> = <em>w</em> = <em>mg</em> = (3.0 kg) (9.8 m/s²) = 29.4 N

The coefficient of static friction between the table and box is 0.40, so the maximum magnitude of static friction is

<em>f</em> = 0.40 <em>n</em> = 0.40 (29.4 N) = 11.76 N

which is to say, the box will not move unless a force larger than this is applied to the box.

10 N is of course smaller than 11.76 N, so the box would stay at rest, and its acceleration would remain 0 m/s².

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I will answer this in English.

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