Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 6.28
A 1.2 kg block has an initial speed of 0.75 m/s as it slides down a ramp at a point 0.40 m above a horizontal floor. Neglect friction and drag. What is the final speed of the block along the horizontal floor?(A) 2.0 m/s.
(B) 2.7 m/s.
(C) 2.8 m/s.
(D) 2.9 m/s.
Correct answer: (D)
Total mechanical energy is conserved, so:
0 = delta(K_tr) + delta(U_grav),
0 = (1/2)*m*(v_f^2 - v_i^2) + m*g*(y_f - y_i),
where y_f = 0, and the common mass m term cancels, such that:
v_f = sqrt(2*g*y_i + v_i^2) = 2.9 m/s.
with v_i = 0.75 m/s, g = 9.80 m/s^2, and y_i = +0.40 m.
Response (C) has v_i = 0; response (B) has a sign error within the square root; and response (A) has both v_i = 0 and a factor error, with v_f = sqrt(g*y_i).
Student responses
Section 72177
(A) : 1 student
(B) : 5 students
(C) : 2 students
(D) : 4 students
Success level: 38%
Discrimination index (Aubrecht & Aubrecht, 1983): 1.00
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