Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 4.26, 4.51
A Physics 205A student pushes an 2.00 kg book across the top of a horizontal table. The coefficient of kinetic friction is 0.3. After it is released, the book slides across the table, and because of friction, slows down with an acceleration of magnitude:
(A) 0.6 m/s2.
(B) 1 m/s2.
(C) 3 m/s2.
(D) 6 m/s2.
Correct answer (highlight to unhide): (C)
The book has two vertical forces acting on it:
Weight force of Earth on book (downwards, magnitude w = m·g = 19.6 N).Because the book is stationary in the vertical direction, these two forces are equal in magnitude and opposite in direction, due to Newton's first law.
Normal force of floor on book (upwards, magnitude N = 19.6 N).
The book has only one horizontal force acting on it (as it moves to the right):
Kinetic friction force of floor on book (to the left).The book is already unstuck and sliding (presumably to the right) after it being released, such that the only horizontal force acting on it is the (constant) kinetic friction force, directed to the left:
fk = µk·m·g = (0.3)(2.00 kg)(9.80 N/kg) = 6 N,
and from Newton's second law, the kinetic friction force (directed to the left) is the sole horizontal force that contributes to the (non-zero) horizontal net force (directed to the left), such that we can solve for the magnitude of the the horizontal acceleration:
ΣFx = –6 N,
m·ax = –6 N,
ax = (–6 N)/m = (–6 N)/2.00 kg) = –3 N/kg = –3 m/s2,
which has a magnitude of 3 m/s2 directed to the left.
(Response (A) is µk·m; response (B) is µk·g/m; and response (D) is the magnitude of the kinetic friction force fk = µk·m·g.)
Student responses
Sections 0906, 0907
(A) : 13 students
(B) : 2 students
(C) : 7 students
(D) : 19 students
Success level: 18%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50
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