A stationary 10 kg object is located on a table near the surface of the earth. The coefficient of kinetic friction between the surfaces is 0.2. A horizontal force of 40 N is applied to the object.

2.04m/s²

Explanation:

Complete Question

A stationary 10 kg object is located on a table near the surface of the earth. The coefficient of kinetic friction between the surfaces is 0.2. A horizontal force of 40 N is applied to the object. Find the acceleration of the object.

According to Newtons second law;

\sum F_x = ma_x

F_m - F_f = ma_x

Fm is the applied force

Ff is the frictional force

m is the mass

a is the acceleration

Substitute the given values

40N - nmg = 10a

40 - 0.2(10)(9.8) = 10a

40 - 19.6 = 10a

20.4 = 10a

a = 20.4/10

a = 2.04m/s²

Hence the acceleration of the object is 2.04m/s²

50 inches should be the resultant vector

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