5.13
(a) We use the workenergy theorem to find the work.
()
2
22
11
1
0
70 kg
4.0 m s
2
i
WK
Em
v m
v
=
∆
=−=
−
=
2
5.6
10 J
−×
.
Assignment 5 solutions (Fall term Physics 102)
(b)
( ) ( ) ( ) ( )
cos
cos180
kk
k
sf
s
n
s
m
g
s
θµ
µ
== °
=
−
=
−
WF
,
so
( )
( )
2
2
5.6
0.70
70 kg
9.80 m s
k
W
s
mg
=−
=
1.2 m
.
5.25
(a) When the rope is horizontal, the swing is 2.0 m above the level of the bottom of the
circular arc, and
( )( )
40 N
2.0 m
g
PE
mgy
==
=
80 J
.
(b) When the rope makes a 30° angle with the vertical, the vertical distance from the
swing to the lowest level in the circular arc is
( )
cos 30
1
cos30
yLL
L
°
=
−
°
and the potential energy is given by
( ) ( )( )( )
1
cos 30
40 N
2.0 m
1
cos 30
g
PE
mgy
mgL
− °
=
=
11 J
.
(c) At the bottom of the circular arc,
y
0
=
. Hence,
g
PE
=
0
.
5.34
At maximum height,
v
and
=0
y
( )
=
40 m s cos60
20 m s
xi
x
vv
=°
=
.
Thus,
20 m s
fx
y
v
=+
=
. Choosing
0
g
PE
=
at the level of the launch point,
conservation of mechanical energy gives
fi
PE
KE
f
=
−
, and the maximum height
reached is
( )
( )
( )
2
40 m s
20 m s
2
29
.80
ms
if
f
y
g
−
−
=
61 m
.
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 Fall '08
 VANDENBERG
 Physics, Energy, Force, Work, WorkEnergy Theorem, Cos, 1.5 m, 1.2 m, 2 2 g

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