p05 071

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71. We mention that the textbook treats this particular arrangement of blocks and pulleys in extensive detail

in Sample Problem 5-5. Using the usual coordinate system (right =

+

x and up =

+

y) for both blocks

has the important consequence that for the 3.0 kg block to have a positive acceleration (a > 0), block
M must have a negative acceleration of the same magnitude (

−a). Thus, applying Newton’s second law

to the two blocks, we have

T

=

(

3.0 kg)



1.0 m/s

2



along x axis

T

− Mg = M



1.0 m/s

2



along y axis .

(a) The first equation yields the tension T = 3.0 N.

(b) The second equation yields the mass M = 3.0/8.8 = 0.34 kg.


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