Question

A 4.40-kg block is set into motion up an inclined plane with an initial speed of Vi = 8.20 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of 0 = 30.0° to the horizontal. (a) For this motion, determine the change in the block's kinetic energy. (b) For this motion, determine the change in potential energy of the block-Earth system. (c) Determine the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

Answer 1

Solution :(a) Change in Kinetic energy, OK: 0-4 mv; - - }(4.4 kg)(8.2 mg ) AK = -147.928 J (b.) Change in potential energy of the block - earth system, AP = mgh = mga cosť = (4.4kg)4.313) (3.com) _ sin ma (30.09) AP = 64.746 J. to work do 10.) Change in Total energy of earth-block system is equal - done by frictional force AK + AP = WA -147.928 + 64.746 Te = 27.73 N ] my coso . (d.) frictional force, fu = Mx FN tu E fy=o. mgsino e FN – mgros 0 = 0 1. Fw = mg coso. Fv = (4.4kq) (9.81 mm) cos( 30.0°) = 37.38 N. as, fx = Mx FN (27.73 N)-( Mix) (37.38 N) MK = 0.742

The coefficient of kinetic friction = 0.742