Question

A block of mass 2kg rests on a plane inclined at 30 with the horizontal the coefficient of friction between the block and the surface is 0.7 calculate the frictional force acting on the block

Answer 1

Mass of the block = 2 kg

Weight of the block = mg = 2 × 9.8 = 19.6 N

The component of the weight along normal = 19.6 Cos 30° = 16.97 N

Hence, the normal force (N) = 16. 97 N

Now, the component of the weight along the inclined plane = 19.6 Sin 30°

Along the inclined plane = 9.8 N

Now, the friction = μ N = 0.7 × 16.97 = 11.87 N

Since, 9.8 N is less than the maximum value of the friction (11.87 N)

Hence, the frictional force = 9.8 N