Physics quiz question: comparing work done by friction in stopping sliding boxes

Physics 205A Quiz 4, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.9

Two boxes of the same mass and initial speed slide across two different horizontal floor surfaces, and each slide different distances to come to a complete stop. The box that slid a shorter distance to stop had __________ work done against it by kinetic friction, compared to the box that slid a longer distance.
(A) less.
(B) the same amount of.
(C) more.
(D) (Not enough information is given.)

Correct answer: (B)

Starting with the energy balance equation:

Wnc = ∆KEtr + ∆PEgrav + ∆PEelas,

where ∆PEgrav = 0 (as there is no change in elevation), ∆PEelas = 0 (as there are no springs), such that:

Wnc = ∆KEtr,

Wnc = (1/2)·m·vf2 – (1/2)·m·v02,

where the final speed v of both boxes is zero; and since both boxes also had the same initial speed v0, then the same amount of (negative) work was done by friction on each box.

Sections 70854, 70855, 73320
Exam code: quiz04w3Rc
(A) : 15 students
(B) : 17 students
(C) : 33 students
(D) : 0 students

Success level: 26%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.49

No comments: