22
1
sin
FTm
g
m
a
Tm
a
θ
−−
=
=
which can be added to obtain
F
–
m
2
g
sin
= (
m
1
+
m
2
)
a
. This yields the acceleration
2
2
12 N (1.0 kg)(9.8 m/s )sin 37
1.53 m/s .
1.0 kg 3.0 kg
a
−°
==
+
Thus, the tension is
T
=
m
1
a
= (3.0 kg)(1.53 m/s
2
) = 4.6 N.
82. We take +
x
uphill for the
m
2
= 1.0 kg box and +
x
rightward for the
m
1
= 3.0 kg box (so
the accelerations of the two boxes have the same magnitude and the same sign). The
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration, Force

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