p05_045.pdf

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Chapter 5 - 5.45
45. The free-body diagram is shown below.
N is the normal force of the
plane on the block and mg is the
force of gravity on the block. We
take the + x direction to be down
the incline, in the direction of the
acceleration, and the + y direc-
tion to be in the direction of the
normal force exerted by the in-
cline on the block. The x com-
ponent of Newton’s second law is
then mg sin θ = ma ;thus,theac-
celeration is a = g sin θ .
.
N
.
.
.
.
.
.
.
•
.
. . . .
.
.
.
.
(+ x )
.
θ
.
. . . .
.
mg
.
(a) Placing the origin at the bottom of the plane, the kinematic equations (Table 2-1) for motion along
the x axis which we will use are v 2 = v 0 +2 ax and v = v 0 + at . The block momentarily stops at its
highest point, where v = 0; according to the second equation, this occurs at time t =
−
v 0 /a .The
position where it stops is
x =
−
1
2
v 0
a
(
=
−
1
2
3 . 50 m / s) 2
(9 . 8m / s 2 ) sin 32 . 0 ◦
−
=
−
1 . 18 m .
(b) The time is
t =
−
v 0
a
=
−
v 0
g sin θ =
−
3 . 50 m / s
(9 . 8m / s 2 ) sin 32 . 0 ◦
−
=0 . 674 s .
(c) That the return-speed is identical to the initial speed is to be expected since there are no dissipative
forces in this problem. In order to prove this, one approach is to set x = 0 and solve x = v 0 t + 2 at 2
for the total time (up and back down) t . The result is
t =
−
2 v 0
a
=
−
2 v 0
g sin θ =
−
3 . 50 m / s)
(9 . 8m / s 2 ) sin 32 . 0 ◦
2(
−
=1 . 35 s .
The velocity when it returns is therefore
v = v 0 + at = v 0 + gt sin θ =
−
3 . 50 + (9 . 8)(1 . 35) sin 32 ◦ =3 . 50 m / s .
.
.
.
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