Physics quiz question: spring-launched box, sliding up ramp

Physics 205A Quiz 4, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.63

A 0.25 kg box is held against a spring that is compressed by 0.10 m. The box is released, and slides up a ramp to a height of 0.70 m above the base of the ramp. Neglect friction and drag. Before the box was released, the elastic potential energy stored in the spring was:
(A) 0.0013 J.
(B) 0.088 J.
(C) 0.25 J.
(D) 1.7 J.

Correct answer (highlight to unhide): (D)

Starting with the the energy balance equation:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where W_nc = 0 (no external gains/losses of mechanical energy) and ∆KE_tr = 0 (as the box is initially held stationary, and again momentarily stationary at its highest point on the ramp) then:

0 = ∆PE_grav + ∆PE_elas,

0 = m·g·∆(y) + (1/2)·k·∆(x²),

0 = m·g·(y_f – y₀) + (1/2)·k·(x_f² – x₀²).

With initial parameters of y₀ = 0 (initial elevation) and x₀ = –0.10 m (compressed from equilibrium), and final parameters y_f = +0.70 m (higher final elevation) and x_f = 0 (spring relaxed after launching the box), then:

0 = m·g·(y_f – 0) + (1/2)·k·(0² – x₀²),

0 = m·g·y_f – (1/2)·k·x₀².

Since the spring strength k is not given, but the mass m of the box is given, the initial elastic potential energy (the second term) cannot be directly calculated, but can be indirectly solved for, as it must be equal to the final gravitational potential energy (the first term):

(1/2)·k·x₀² = m·g·y_f,

(1/2)·k·x₀² = (0.25 kg)·(9.80 m/s²)·(+0.70 m) = 1.715 J,

or to two significant figures, the elastic potential energy that was stored in the spring was 1.7 J.

(Response (A) is (1/2)·m·(x_i)², response (B) is (1/2)·m·y_f, response (C) is m·g·x₀.)

Sections 70854, 70855
Exam code: quiz04iMpL
(A) : 1 student
(B) : 8 students
(C) : 12 students
(D) : 30 students

Success level: 59%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.73