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Serga [27]
2 years ago
11

A 5.0\, \text {kg}5.0kg5, point, 0, start text, k, g, end text box is at rest on a table. The static friction coefficient \mu_sμ

Physics
2 answers:
STALIN [3.7K]2 years ago
5 0

Answer:

0 m/s^2

Explanation:

khan academy

oksian1 [2.3K]2 years ago
4 0

Answer:

25 N

Explanation:

When a force is applied to an object, for it to move it has to overcome the frictional force. The frictional force is perpendicular to the surface on which it acts. For an object at rest, it is acted upon by static friction, this friction must be overcome before the object can move, while a moving object is resisted by kinetic friction that prevents it from moving.

Since the box is at rest, it is acted upon by a static friction.

mass = 5 kg, acceleration due to gravity = 10 m/s², static friction coefficient = 0.5

Weight of the box = mass * acceleration due to gravity = 5 * 10 = 50 N

The applied horizontal force (F) is:

F=\mu_s*weight=0.5*50\\\\F=25\ N

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