A package with a mass of 36 kg is being pulled by a force of 215 N parallel to a horizontal surface. The coefficient of kinetic friction between the package and the surface is 0.35. If the package moves a distance of 9.1 m, what is the resultant work?

To calculate the resultant work, find the work done by the applied force and the work done by friction. The resultant work is the sum of these two values. In this case, the resultant work is 830.792 J.

Explanation:

To calculate the resultant work, we need to consider two forces: the applied force and the frictional force. The work done by the applied force is given by Wapplied = Fapplied * d, where Fapplied is the applied force and d is the distance the package moves. In this case, Fapplied = 215 N and d = 9.1 m. Therefore, Wapplied = 215 N * 9.1 m = 1956.5 J.

The work done by friction, Wfriction, is given by Wfriction = -frictional force * d, where the frictional force is equal to the coefficient of kinetic friction times the normal force. The normal force is equal to the weight of the package, which is given by mass * acceleration due to gravity. In this case, mass = 36 kg and acceleration due to gravity ≈ 9.8 m/s². Therefore, the normal force is equal to 36 kg * 9.8 m/s² = 352.8 N. Substituting the values, Wfriction = -0.35 * 352.8 N * 9.1 m = -1125.708 J.

The resultant work is the sum of the work done by the applied force and the work done by friction. Therefore, the resultant work is Wresultant = Wapplied + Wfriction = 1956.5 J + (-1125.708 J) = 830.792 J.