The mass of the two blocks are 4 kg and 20 kg, respectively. The kinetic frictio
ID: 2991702 • Letter: T
Question
The mass of the two blocks are 4 kg and 20 kg, respectively. The kinetic frictional coefficient between the 4kg block and the table is mu k= 0.4. The system is released from rest and the system will move. Gravity acts downward. find the velocity of the 20 kg block when it travels down 2m; if we replace the 20 kg block by a block with mass M, and the static frictional coefficient between the 4kg block and the table is mu s= 0.5, what is the minimum value of M that will cause the system to move once released from rest?Explanation / Answer
a) Applying conservation of energy
change in potential energy = change in kinetic energy + frictional loss
=>20*g*2 = 0.5*20*v2 + 0.5*4*v2 + 0.4*4*g*2 {velocity of both masses will be same}
=>v = 5.48 m/s
b) Static friction = Mg for the system to move
=>0.5*4*g = M*9.8
=>M = 2kg (minimum value of M)
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