Physics final exam question: box sliding down inclined board

Physics 5A Final Exam, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problem 4.146

[20 points.] The coefficient of static friction between a 2.50 kg box and a wooden board is 0.45 and the coefficient of kinetic friction between the box and the board is 0.28. The box is placed on the board, which is slowly lifted at one end until the box starts to slide down the board, at which point the angle theta of the board is held constant. Determine the angle theta of the board, and the acceleration of the box as it slides down the board. Show your work and explain your reasoning.


Solution and grading rubric:

• p = 20/20:
  Correct. At the angle where static friction is (about to be) overwhelmed, it is equal to the x-component of weight, and Newton's first law is used to determine that the angle theta = 24.2 degrees. At this angle, kinetic friction acts on box, and Newton's second law is used to find that the acceleration of the box is 1.52 m/s^2 down along the ramp.
  
• r = 16/20:
  Nearly correct, but includes minor math errors. Correct angle for ramp, but acceleration has minor problems, such as kinetic friction being the only force acting along the x-direction in Newton's second law, or divides by g instead of by mass on both sides of Newton's second law.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Angle of board is wrong, but accounts for both kinetic friction and the x-component of weight acting on the box in Newton's second law. Or determines correct angle of board, but nothing much further than that.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some systematic attempt at both the (limiting) static and accelerating case.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
p: 3 students
r: 4 students
t: 9 students
v: 9 students
x: 7 students
y: 1 student
z: 0 students

A sample of a "p" response (from student 6154) is shown below:
An "r" response (from student 2012), where the sine and cosine components were inadvertently switched, but following results were essentially correct:
Another "r" response (from student 5711), where at the last step, g is divided from both sides of Newton's second law, instead of m:
A "t" response (from student 2880), where only the angle of the board is found: