8 .
Careful—this one’s tricky.
The Answers
(A Step-by-Step Solution to #1 Is on the Next Page.)
1 . a = 6.3 m/s^2 , down the plane.
t = 2.5 s
2 . a = 4.9 m/s^2 , down the plane.
t = 2.9 s
3 . a = 5.2 m/s^2 , down the plane.
t = 2.8 s
4 . a = 4.4 m/s^2 , down the plane.
t = 3.0 s
5 . Here the angle of the plane is 27° by trigonometry, and the distance along the plane is 22 m.a = 4.4
m/s^2 , down the plane.
t = 3.2 s
6 . a = 6.3 m/s^2 , down the plane.
t = 1.8 s
7 . a = 6.3 m/s^2 , down the plane.
t = 3.5 s
8 . This one is complicated. Since the direction of the friction force changes depending on whether
the block is sliding up or down the plane, the block’s acceleration is NOT constant throughout the
whole problem. So, unlike problem #7, this one can’t be solved in a single step. Instead, in order to
use kinematics equations, you must break this problem up into two parts: up the plane and down the
plane. During each of these individual parts, the acceleration is constant, so the kinematics
equations are valid.
• up the plane:
a = 6.8 m/s^2 , down the plane.
t = 0.4 s before the block turns around to come down the plane.
• down the plane:
