A 5.00-kg package slides 1.50 m down a long amp that is inclined at 24.0° below the horizontal. The coefficient of kinetic friction between the package and the ramp is ???k = 0.310 Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after sliding 1.50 m down the ramp?

Solution 68P Mass of the package is 5 kg. The length of the inclined plane is 1.5 meters. 0 The angle of inclination is 24 . Coefficient of kinetic friction is, k = 0.310. a) The frictional force is, F fric= kN = 0.310 × mg = 0.310 × 5 × 9.8 = 15.19 Newton The direction of frictional force is antiparallel to the direction of displacement. So, the work done by the frictional force is, 0 W fric F fricdS = F fric× S × cos 180 = 15.19 × 1.5 × ( 1) = 22.785 jules. The minus sign indicates the work done is negative. b) The gravitational force is acting in the downward direction. F = mg = 5 × 9.8 = 49 N . gravity The displacement is 1.5 meters. The angle between the force and displacement vector is, = 180 (90 + 24) = 180 114 = 66 . The work done by the gravitational force is, W = F. dS = | || |s gravity 0...