A block of weight W is sliding down an inclined plane. The angle of the inclined is represented by 0.The block is lubricated by a thin film of oil with a linear velocity profile and a thickness h. The film contact area is represented by A. First derive an expression for the terminal velocity (acceleration is zero) of the block V. Now if you assume that the mass of the block is 6kg, the contact area is 35 cm, and 0=15. The oil film is SAE oil at 20C and is 1mm thick. What is the block's terminal velocity V? W Liquid film of thickness h Block contact area A