Cuesta College, San Luis Obispo, CA
A "BASE[*]" jumper (total mass of 88 kg with equipment) reached a final terminal speed of 53 m/s after falling 452 meters downwards (before using his parachute)[**]. Drag is not negligible, assume a zero initial speed, and that he dropped straight down. For this process, the decrease in gravitational potential energy of the jumper was __________ the increase in translational kinetic energy of the jumper.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)
[*] Jumping off of a "building, antenna, span, or Earth (i.e., cliff)."
Correct answer (highlight to unhide): (C)
The energy transfer-balance equation is given by:
Wnc = ∆KEtr + ∆PEgrav + ∆PEelas,
where ∆PEelas = 0, as there is no spring involved in this process. Note that on the left-hand side of this equation, the non-conservative work done by the non-neglible drag force against the BASE jumper is:
Wnc = Fdrag·s·cosθ,
and must have a negative value, as the tail-to-tail angle between the upwards drag force and the downwards displacement is θ = 180°, and cos(180°) = –1. (Note that the weight force on the BASE jumper is a conservative force, and is already included in the ∆PEgrav term on the right-hand side of the equation, so the work done by this force is not explicitly considered here.)
Since the left-hand side of the energy transfer-balance equation is negative, then:
(–) = ∆KEtr + ∆PEgrav,
such that decrease in gravitational potential energy (∆PEgrav is negative) must be greater than the increase in translational kinetic energy (∆KEtr is positive).
Sections 70854, 70855, 73320
Exam code: quiz04th1R
(A) : 13 students
(B) : 16 students
(C) : 22 students
(D) : 0 students
Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.82