How Do You Solve a Kinetic Friction Problem Involving Two Blocks?

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SUMMARY

The discussion centers on solving a kinetic friction problem involving two blocks, where the lower block is subjected to a tension force of 20 N and has a coefficient of kinetic friction of 0.30 with both the surface and the upper block. The initial attempt at calculating the acceleration of the 2.0 kg block yielded an incorrect result of 2.75 m/s². After correcting the equations to account for friction between the blocks, the correct acceleration was determined to be 1.77 m/s².

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Homework Statement


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.
knight_Figure_08_28.jpg


Find the acceleration of the 2.0kg block

Homework Equations


The Attempt at a Solution

I set m1a= T - (0.3)m1g and m2a= 20 - T - 0.3m1+2g. After which I added them and isolating a, giving me a=(20 - 0.3m1+2g - 0.3m1g)/(m1+m2), which gave me the answer 2.75, but apparently it is not correct.

Never mind, I solved it. I forgot the friction between block 1 and 2 for movement of block 2
 
Last edited:
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m1a= T - (0.3)m1g
m2a= 20 - T - 0.3m1+2g-0.3m1g

==>a=[20-(m1+m2)g*0.3-2*0.3*m1g]/(m1+m2)=5/3 (ms^-2)

is it right?
 
Yep, you got the equation set up right, but the answer is 1.77, not 5/3
 

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