1. The problem statement, all variables and given/known data A man pulls a 15-kg sled containing a 5.0-kg dog along a straight path on a horizontal surface. He exerts a force of 55 N on the sled at an angle of 20.° above the horizontal, as shown in the figure above. The coefficient of friction between the sled and the surface is 0.22. a. Create a free body diagram for the forces on the sled b. Calculate the normal force acting on the sled c. Calculate the work done by the pulling force if the sled is moved 7m d. At some later time, the dog rolls off the side fo the sled. The man continues to pull with the same force. Sketch a graph of KE versus time t for the sled. Include both the sled's travel with and without the dog on the sled. Clearly indicate with the symbol tr the time at which the dog rolls off. 2. Relevant equations W=FL W=cos@Fl Fn=mg-Fpullsin@ Ff=u(mg-Fpullsin@) 3. The attempt at a solution My thought is that the 55 N force is acting only on the 15kg of the sled, not the 5kg mass or a 20kg system. Therefore, the free motion diagram will have a 147 N (15*9.8) force of gravity acting on the sled, the 55 N force acting on a 20 degree angle, the normal force of the sled (147-Fpullsin@=147-19=128 N), and the force of friction, which does include the 5kg dog, which would be 0.22(20kg*9.8-55sin20)=38.6 N. The normal force on the sled would be 15kg*9.8-55sin20=128 N The work done by the pulling force on the sled is equal to the horizontal component of that force times the distance: (55cos20*7.0m)=361.8 N I am not sure how specific I need to make the graph, but it would appear to me that the Net force acting on the sled in the horizontal direction would increase from 13 N to 24 N before and after the dog roles off, respectively, because the horizontal component of force remains constant at 52 N but the resisting force of friction decreases from 39 N to 28 N. Therefore, the acceleration of the sled would increase from (13.1=15a) a=0.87m/s^2 to (23.54=15a) a=1.57m/s^2, so its kinetic energy, holding mass constant at 15kg, would be increasing by an exponential amount both before and after the dog roles off, but would increase even faster after this event because KE=mv^2/2, and if a is positive, v must be increasing linearly, which would translate to an exponential increase in kinetic energy through this equation. Is this right, or am I completely off base in my thinking? Can the graph of kinetic energy as a function of L be calculated exactly, or should I create a more general sketch?