Physics midterm question: net force of two opposite forces on accelerating object

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A physics question on an online discussion board[*] was asked and answered:

P-dog: If there are only two opposing forces acting on an accelerating object, is it possible for the net force to ever be larger than either of those two forces?
OSG: No.

Discuss why this answer is correct, and how you know this. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

[*] answers.yahoo.com/question/index?qid=20180916184605AASRFkI.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. the net force is the vector addition of all forces (in this case, only two forces) acting on the same object; and
  2. since these two forces are opposite in direction, one force must be larger than the other in order for the object to be accelerating (Newton's second law); then
  3. the net force cannot be more than the larger of the two forces acting on the object, as the net force is the difference of the magnitudes of these two forces (i.e., the magnitude of the larger force minus the smaller magnitude force). (It is also possible that the net force could be smaller than both of the two forces acting on the object.)
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's laws to a free-body diagram.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws to the forces on a free-body diagram.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 33 students
r: 11 students
t: 2 students
v: 10 students
x: 1 student
y: 0 students
z: 1 student

A sample "p" response (from student 1842):
Another sample "p" response (from student 1996):