Answer:
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
