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The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.30. The coefficient of kinetic friction between the lower block and the upper block is also 0.30. What is the acceleration of the 2.0 kg block??
Forces of A: T(.3*1kg*9.8m/s/s)= (1kg)a
Forces on B: 20N  T (.3 * 3kg * 9.8m/s/s)= (2kg)a
If you Add them together: 20N(.9*9.8)(.3*9.8) = 3a
20N11.76 = 3a
8.24/3 = a = 2.74667m/s/s
Should it be 3kg as the mass used in finding the kinetic friction force in the formula for the forces on B or should it be 2kg? And should it be 2kg as the mass on the right side of the equation for the forces on B or should it be 3kg????
the answer of 2.75 is not correct. What am I doing wrong???
Thank you.
Stephen Doty
Forces of A: T(.3*1kg*9.8m/s/s)= (1kg)a
Forces on B: 20N  T (.3 * 3kg * 9.8m/s/s)= (2kg)a
If you Add them together: 20N(.9*9.8)(.3*9.8) = 3a
20N11.76 = 3a
8.24/3 = a = 2.74667m/s/s
Should it be 3kg as the mass used in finding the kinetic friction force in the formula for the forces on B or should it be 2kg? And should it be 2kg as the mass on the right side of the equation for the forces on B or should it be 3kg????
the answer of 2.75 is not correct. What am I doing wrong???
Thank you.
Stephen Doty
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