A block is released from rest at the top of an inclined plane making an angle of 37^ with the horizontal. The first 2 m of the incline is perfectly smooth, while the remaining 6 m is rough. If the block comes to rest exactly at the bottom of the rough section, the coefficient of kinetic friction of the rough section is: (Take 37^ = 3 5 and 37^ = 4 5 )

Step-by-step solution

Using the work-energy theorem, the total work done on the block is equal to its change in kinetic energy. Since the block starts from rest and finally comes to rest, K = 0 . The forces doing work are gravity (over the entire length L₁ + L₂ ) and kinetic friction (only over the rough section L₂ ). W_ gravity + W_ friction = 0 mg(L₁ + L₂) - _k mg L₂ = 0 (L₁ + L₂) = _k L₂ _k = L₁ + L₂ L₂ Given L₁ = 2 m, L₂ = 6 m, and = 37^ : _k = 2 + 6 6 37^ _k = 8 6 3 4 _k = 4 3 3 4 = 1.0 Answer: 1.0