A block of mass 1.5 kg is placed on a rough table (horizontal) and is pulled by a constant force of 1.2 kgf. The coefficient of friction between...

Hello!

By Newton's Second law, the acceleration is equal to the net force divided by the mass of a body. The net force is the vector sum of all forces acting on a body.

These forces in our problem are the traction force (horizontal), the friction force (horizontal), the gravity force mg and the reaction force R (both vertical).

Because the movement is horizontal, the acceleration and the net force are also horizontal. So the...

Hello!

By Newton's Second law, the acceleration is equal to the net force divided by the mass of a body. The net force is the vector sum of all forces acting on a body.

These forces in our problem are the traction force (horizontal), the friction force (horizontal), the gravity force mg and the reaction force R (both vertical).

Because the movement is horizontal, the acceleration and the net force are also horizontal. So the vertical forces are balanced, R = mg. Next, the friction force is 0.3R = 0.3mg = 0.45g (in Newtons).

The traction force is 1.2 kgf = (1.2g) Newtons. Thus the net force is equal to 1.2 g - 0.45 g = 0.75 g (N), and the acceleration is (0.75 g)/1.5 = 0.5 g.

So the answer in terms of g is 0.5.