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A 8 kg block is being pulled horizontally on a table by a 6 kg block that hangs in the air by a cord fasten to the 8 kg block over a pulley, there is only friction between the table and the 8 kg block. Both blocks are initially moving with 0,900 m/s and they come to rest after 2 meters.
I have to use the work-energy theorem to find the coefficient of kinetic friction.
What I've done so far is:
total work done on the 6 kg block must be -0.5*6*0.9*0.9 = 2.43 J
So the net force must be -2.43J/2m = -1,215N
Fy=-mg+T=-1,215 so the tension is 57.58 N
Total work done on the 8 kg block must be -0.5*8*0.9*0.9 = -3.24 J
So the net force on the block must be
Fx=57.58-fk=-3.24J/2m
coefficient of friction must be 55.96/8*9.8=0.714
I don't get the right answer which is 0.786
Could someone please give me a hint to what I am doing wrong.
With thanks in advance
I have to use the work-energy theorem to find the coefficient of kinetic friction.
What I've done so far is:
total work done on the 6 kg block must be -0.5*6*0.9*0.9 = 2.43 J
So the net force must be -2.43J/2m = -1,215N
Fy=-mg+T=-1,215 so the tension is 57.58 N
Total work done on the 8 kg block must be -0.5*8*0.9*0.9 = -3.24 J
So the net force on the block must be
Fx=57.58-fk=-3.24J/2m
coefficient of friction must be 55.96/8*9.8=0.714
I don't get the right answer which is 0.786
Could someone please give me a hint to what I am doing wrong.
With thanks in advance