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 the block and the surface is 0.3. Find the acceleration produced in terms of 'g'.

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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...

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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.

 

 

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