Cuesta College, San Luis Obispo, CA
Martin Howey
youtu.be/gATHEjz_vMk
An athlete drags a set of tires a distance of 100 m to the right across the ground at a constant speed, pulling 360 N on a strap that makes an angle of 20° above the horizontal[*]. Friction is not negligible. For this process, the work done by the athlete on the tires has a magnitude of:
(A) 0 J.
(B) 1.2×104 J.
(C) 3.4×104 J.
(D) 3.6×104 J.
[*] youtu.be/gATHEjz_vMk.
Correct answer (highlight to unhide): (C)
The (non-conservative) work done by the athlete on the tires is given by:
Wstrap = (Fstrap·cosθ)·s,
where the tail-to-tail angle between the force of the strap on the tires and displacement of the tires is θ = 20°, such that:
Wstrap = (360 N)·cos(20°)·(100 m) = +33,828.93 J,
or to two significant figures, the magnitude of the work done on the tires by the athlete pulling on the strap is 3.4×104 J.
(Response (A) is the net external work done on the tires, as there is no change in translational kinetic energy (constant speed), no change in gravitational potential energy (no change in elevation), and no change in elastic potential energy (no springs), such that the positive work done by athlete on the tires must be equal in magnitude to the negative work done by the kinetic friction force on the tires. Response (B) is Fstrap·s·sin(20°); response (D) is Fstrap·s.)
Sections 70854, 70855
Exam code: quiz04w33N
(A) : 0 students
(B) : 2 students
(C) : 33 students
(D) : 13 students
Success level: 69%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.58
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