Question
When the three blocks inFig. 6-29 are released from rest, theyaccelerate with a magnitude of 0.500m/s2. Block 1 has mass M, block 2has 2M, and block 3 has 2M. What isthe coefficient of kinetic friction between block 2 and the table?
Step 1
For block 1, the forces acting on it are the tension T1 in the direction of acceleration, and its weight Mg acting downwards. According to Newton's second law, we can write the equation for block 1 as T1 - Mg = Ma. Rearranging this equation gives us T1 = Ma + Mg. Show more…
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When the three blocks in Fig. $6-29$ are released from rest, they accelerate with a magnitude of 0.500 $\mathrm{m} / \mathrm{s}^{2} .$ Block 1 has mass $M,$ block 2 has $2 M,$ and block 3 has 2$M$ . What is the coefficient of kinetic friction between block 2 and the table?
If the coefficient of kinetic friction between the block and the table in Fig. 4.40 is 0.560 , and $m_{1}=0.150 \mathrm{~kg}$ and $m_{2}=0.250 \mathrm{~kg},$ (a) what should $m_{3}$ be if the system is to move with a constant speed? (b) If $m_{3}=0.100 \mathrm{~kg}$, what is the magnitude of the acceleration of the system?
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