Prof. swarzenneger picks up his 45 kg brief case on the 17th floor of his office building (85 m above the ground) and prepares to go to class. he walks 24 meters straight to the elevator at 1.5 m/s, takes the elevator to the ground floor at a steady 3 m/s, and walks 175 m in a straight line at 2 m/s south to the classroom, where he deposits his briefcase on the floor. about how much work has he done on the briefcase over the entire trip?

The work done by the professor on the briefcase is zero.

Explanation:

Hi there!

Work is done by a force on an object when the applied force (or one of its components) is parallel to the direction of movement.

The professor applies a force in the briefcase that is perpendicular to the displacement (it points in the vertical direction) that allows him to keep the briefcase in the air. When the professor enters the elevator, it is the elevator that moves him along with the briefcase, so there is no work done by the professor there.

During all the trajectory, the professor did not move the briefcase in the vertical direction except when he picks it up and then deposits it on the floor of the classroom.

When he picks the briefcase, the work done on it is positive (the force is applied in the same direction as the displacement) but when he deposits it in the floor, the work done is negative (the force is applied in opposite direction to the displacement of the briefcase). When he picks the briefcase up, the magnitude of the work done by the professor has to be greater than when he deposits it because the gravity force does work in the down direction that helps him to deposit the briefcase. We can assume that the distance of the briefcase to the floor is small and thus neglect this work.

Then, the net work done by the professor on the briefcase is zero.

c. magnitude and direction