Question

A block of mass m = 12 kg lies on a rough inclined plane with a coefficient of kinetic friction μc = 0.8. block position is (35; 28) m. Furthermore, F forms an angle α with the inclined plane (α = θ). If the modulus of F equals 126,205 N. a) Found the modulus of the normal floor in the block. b) the acceleration module.

Answer 1

Answer:

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