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
s

mu, start subscript, s, end subscript between the box and table is 0.500.500, point, 50, and the kinetic friction coefficient \mu_kμ
k

mu, start subscript, k, end subscript is 0.300.300, point, 30. Then, a 20\,\text N20N20, start text, N, end text horizontal force is applied to the box.
What is the best estimate of the magnitude of the box’s acceleration?

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:

[tex]F=\mu_s*weight=0.5*50\\\\F=25\ N[/tex]

smangar smangar · Answer 2 · 2021-02-25T17:19:13+01:00

smangar smangar

25-02-2021

Answer:

0 m/s^2

Explanation:

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