the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0º slope at constant speed, as shown in Figure 7.37. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
Step-by-step solution Step 1 of 11 Work done by a force over a displacement can be calculated using the following relation: Here F and d are the magnitude of force and displacement respectively. The angle between force and displacement is represented by . Consider the following free body diagram which represents the sled-victim problem given: Step 2 of 11 (a) Work done by frictional force will be given by: Here is the frictional force and d is the displacement. The angle between them is . Step 3 of 11 The frictional force can be found using the following formula: Here is the coefficient of friction and N is the normal force on the sled and the victim. From the figure, it can be seen that N is equal to . So, the frictional force will be: Step 4 of 11 Then, the work done by the frictional force will become: Substitute 90.0 kg for m, 30.0 m for d, for g, 0.100 for , 180° for and 60° for . Therefore, work done by the frictional force is . Step 5 of 11 (b) The work done by the rope will be: Here T is the tension generated in the rope. Step 6 of 11 The free body diagram shows that: