An 8.00-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0° below the horizontal. The coefficient of kinetic friction between the package and the chute’s surface is 0.40. Calculate the work done on the package by(a) friction (b) gravity, and (c) the normal force (d) What is the net work done on the package)
Solution 10E Introduction We will draw a figure of the situation containing all the forces, then we will identify each forces and what are the angles with the displacement, then we will calculate the total work done on the box. Step 1 The work done by a force is given by W = Fdcos Where F is the force d is the displacement and is the angle between force and displacement. Step 2 The following figure shows all the forces that is acting on the box. Step 3 (a) The frictional force is given by F f N Where = 0.40 is the coefficient of friction and N is the normal reaction. The normal reaction is the reaction of the force applied by the object to the surface. And from the figure we can see that the force applied by the box to the surface is mg cos So magnitude of normal reaction is N = mg cos Hence the frictional force is 2 F f N = mg cos = (0.40)(8.00 kg)(9.8 m/s )cos(53.0°) = 18.87 N The frictional force always act against the velocity, or direction of motion. Hence the angle between the frictional force and the displacement is 180°. Also, since the length of the chute is 2.00 m, the magnitude of displacement is d = 2.00 m Hence the work done by frictional force is W = F dcfs = (18.87 N)(2.00 m)cos(180) = 37.7 J Hence the work done by the friction on the box is -37.7 J.