20131031

Physics quiz question: airplane trading altitude for speed

Physics 205A Quiz 4, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 6.79

"crop duster"
cdn-pix
flic.kr/p/8ciFEK

A crop duster airplane[*] is flying at its cruise speed of 64 m/s (143 mph). It descends 20 m in order to "trade altitude for speed."[**] Ignore friction/drag, and work done by the engine during this process. The final speed of the airplane after its descent is:
(A) 20 m/s.
(B) 67 m/s.
(C) 84 m/s.
(D) 110 m/s.

[*] wiki.pe/Air_Tractor_AT-400.
[**] Planes, Walt Disney Pictures (2013).

Correct answer (highlight to unhide): (B)

The energy transfer-balance equation is given by:

Wnc = ∆KEtr + ∆PEgrav + ∆PEelas,

where Wnc = 0 (no external gains/losses of mechanical energy due to the engine or friction/drag), and ∆PEelas = 0 (no springs involved), such that the remaining terms in the equation are:

0 = ∆KEtr + ∆PEgrav,

0 = (1/2)·m·(vf2v02) + m·g·(yfy0).

The mass m cancels out, and solving for the final speed vf:

g·(yfy0) = (1/2)·(vf2v02),

v02 – 2·g·(yfy0) = vf2,

√(v02 – 2·g·(yfy0)) = vf,

√((64 m/s)2 – 2·(9.80 m/s2)·((0 m) – (20 m))) = vf,

then vf = 66.9925369 m/s, or to two significant figures, 67 m/s.

(Response (A) is √(–2·g·(yfy0)); response (C) is v0 + √(–2·g·(yfy0)); response (D) is √(–v0·g·(yfy0)).)

Sections 70854, 70855, 73320
Exam code: quiz04iSs5
(A) : 0 students
(B) : 23 students
(C) : 41 students
(D) : 2 students

Success level: 35%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.66

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