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In the apparatus shown in the figure (Figure 1) , m1 = 10kg and the coefficients of static and kinetic friction between m1 and the table are 0.69 and 0.32, respectively.


Part A
What mass of m2 will just barely set the system in motion?

1 answer

W1 = m * g = 10kg * 9.8N/kg = 98 N. = Wt. 0f m1.

Fs = u * mg = 0.69 * 98 = 67.62 N. = Force of static friction.

m2 * g = Fs = 67.62 N.
m2 * 9.8 = 67.62
m2 = 6.9 kg.
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