A metal block of mass 395 kg slides at constant velocity V m

A metal block of mass 39.5 kg, slides at constant velocity V m/s,down a planar surface that is inclined at an angle of 7.2 degrees to the horizontal.The planar surface is covered with film of oil that is 0.1 mm thick.The surface of the block that is in contact with the oil has an area of 0.46 m2 and the viscosity of the oil is 0.071 PA s .Assuming that Assuming that the magnitude of the shear stress on the surface on the metal block that is in contact with the oil is T N/m2 ,what is the velocity V of the block.

Solution

velocity of the block is calculated as follows:

    v = mgysin@/Au

       = 39.5*9.8*0.1*10^-3*sin7.2 / 0.46*0.071

       = 0.1485 m/s