Cuesta College, San Luis Obispo, CA

(A) gravitational potential energy.
(B) translational kinetic energy.
(C) (Both of the above choices.)
(D) (Neither of the above choices.)
Correct answer (highlight to unhide): (A)
The energy transfer-balance equation is given by:
Wnc = ∆KEtr + ∆PEgrav + ∆PEelas,
where ∆PEelas = 0, as there is no spring involved in this process.
Just looking at the two remaining terms on the right-hand side of the energy transfer-balance equation, for the change in translational kinetic energy:
∆KEtr = (1/2)·m·(vf2 – v02),
and since the speed is constant, v0 and vf have the same magnitude, then KEtr is constant (∆KEtr = 0).
Also for the change in gravitational potential energy:
∆PEgrav = m·g·(yf – y0),
and since yf is greater than y0, then PEgrav increases (∆PEgrav is positive).
(In order for the equality to hold for the energy transfer-balance equation, the student must then be doing positive work on the box in order to increase its gravitational potential energy.)
Sections 70854, 70855
Exam code: quiz04W3rK
(A) : 39 students
(B) : 3 students
(C) : 5 students
(D) : 4 students
Success level: 76%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.48
No comments:
Post a Comment