Physics Quiz Answers and Solution

During a physics experiment on earth at sea level but in a room that has all of its air evacuated, a mass is accelerating upward while being pulled by a string. Besides the string, nothing touches the mass. Below you see a diagram that only shows the string force acting on the mass. In the situation shown, the acceleration is 5m/s2 upward, and FS is the force due to the string. The mass is 600g.This was version A.

Version B had a = 5m/s2 and M = 2700 g. In Version B, #2’s solution yieldsFS = 40.5 N

Version C had a = 20 m/s2 and M = 450 g.In Version C, #2’s solution yieldsFS = 13.5 N

Version D had a = 20 m/s2 and M = 100 g. In Version D, #2’s solution yieldsFS = 3N

FS = ?

  1. Above was an effort to make a free-body-diagram of the forces that act on the mass. Complete the free-body diagram correctly if it needs any additional forces by adding such forces using an arrow or arrows in the proper direction(s). Any vector arrow you make must be labeled with a true label. (Such a label can either be symbolic or a correct numerical value.) If you say that the FBD above is complete, just write the word “complete” and don’t do anything to the diagram.

A downward vector that says g gets zero credit. A person who objects and says “but g and gravity” are the same thing has not read the multiple places in my solution notes where I’ve clearly shown why g and gravity are not the same thing, and they also didn’t think it important enough when I said it in class.

A downward label that says any of the following gets full credit: Mg, “gravity”, Fgrav, Fg, W, Weight. All of those letter signals just listed have the exact same numerical value, they are all synonymous with each other by definition and none of them equal g. Speaking of numerical values, another way to get full credit is to put a label with a number of N equal to whatever Mis times g.But theMhad to be in kg. One more way to get full credit: Use M in g, and write the value (600 g)(10 m/s2) = 6000 milliNewton

  1. Use Newton’s 2nd Law (F = Ma) to calculate the value of the string force FS.

Calculating F = Ma = (0.6 kg)(5 m/s2) = 3 N gets less than full credit. Ma leads to an answer for Fnet. The question asks for FS.FS≠ Fnet. The solution is:

Ma = FS–Mg → FS = Ma + Mg = (0.6 kg)(5 m/s2) + (0.6 kg)(10 m/s2) = 9 N = FS

Or

Knowing that Ma = (0.6 kg)(5 m/s2) = 3 N, I say 3 Nis definitely thenet force, and how can that happen in the presence of 6 N of downward gravity? It happens only if the upward force is 9 N, because 9 is 3 greater than 6. The FS has to be at least as strong as gravity to have no downward acceleration, and it’s 3 N greater since Fnet = +3 N.