A block of mass 5kg is on a rough slope inclined at an angle of 30 degrees to the horizontal, it is at the point of sliding down the slope. Calculate the coefficient of friction between the block and the slope.

If the block is at the point of sliding, we know the frictional force, F, is at it's maximuim, F = kR, where k is the coefficient of friction and R is the reaction force.As we know the mass of the block, the weight of the block is 5g.We can then resolve the weight into components parallel and perpendicular to the slope.The component parallel to the slope = 5g sin(30) by trigonometry.The component perpendicular to the slope = 5g cos(30)The block is at the point of sliding, so it is currently at rest.We therefore know by newtons first law, that the resultant force is zero.So the magnitude of the frictional force must be equal to the component of weight parallel to the slope, and the magnitude of the reaction force must be equal to the perpendicular component of the weight.So we getF = 5g sin30R = 5g cos30so rearranging F = kRk = (5g sin30)/(5g cos30) = sin 30/cos 30sin/cos is just tan! so we getk = tan30 = 1/sqrt(3) which is approximately 0.577 ( 3 sig fig)