Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.13
Mundo Ferroviário
youtu.be/FmVSUl3cI5U
A 850,000 kg passenger train traveling at 4.2 m/s on a horizontal track pushes up against an end-of-track buffer stop[*]. As a result, the buffer stop brings the passenger train to a complete rest after traveling 7.0 m. Neglect friction and drag. The magnitude of the average force exerted by the buffer stop on the train is:
(A) 2.6×105 N.
(B) 1.1×106 N.
(C) 2.1×106 N.
(D) 5.8×107 N.
[*] "The buffer stop is designed to move up to 7 m to slow down a 850 t passenger train from 15 km/h," wiki.pe/File:Buffer_stop_zurich.jpg.
Correct answer (highlight to unhide): (B)
The non-conservative work done by the buffer stop (considered as an external agent outside of the train's energy systems) decreases the train's translational kinetic energy:
Wnc = ∆KEtr,
where the work done is the product of the average force exerted and the displacement:
Wnc = Fav·s·cosθ,
such that the average force is:
Fav·s·cosθ = ∆KEtr,
Fav = ∆KEtr/(s·cosθ),
Fav = (1/2)·m·(vf2 – v02)/(s·cosθ),
Fav = = (1/2)·(850,000 kg)·((0 m/s)2 – (4.2 m/s)2)/((7.0 m)·cos(180°)) = 1,071,000 N,
or to two significant figures, the magnitude of the force is 1.1×106 N.
(Response (A) is (1/2)·m·v0/(2·s); response (C) is m·v02/s; response (D) is m·g·s.)
Sections 70854, 70855
Exam code: quiz04Bmw3
(A) : 7 students
(B) : 20 students
(C) : 17 students
(D) : 7 students
Success level: 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.80
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