Two blocks on top of each other accelerating

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The discussion focuses on calculating the acceleration and time required for two blocks, one on top of the other, to move without the top block sliding off. The static friction coefficient for the top block is 0.60, while the kinetic friction coefficient for the bottom block is 0.20. The initial calculations incorrectly assumed the driving force acted on both blocks, leading to an erroneous acceleration of -9.8 m/s². The correct approach identifies that the static friction from the top block is the driving force for the bottom block, necessitating a revision of the equations used to determine the motion.

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FancyNut
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I did this question several times and all I get for acceleration is negative 9.8...

static friction for the top block is .60 and kinetic friction for the bottom block is .20... The force is making both blocks move a distance of 5 meters starting from rest. They want to know What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

These are the equations I used...

acceleration is equal for both so a_t = a_b = a.

for the top block:

\sum F_x = F = m_t a + f_s

F = m_t a + u_s m_1 g

for the bottom block:

\sum F_x = F = m_b a + f_k

F = m_b a + u_k g (m_t + m_b)


I subtracted second equation from the first and get:

0 = a (m_t - m_b) + u_s m_1 g - u_k g (m_t + m_b)

a = - 9.8


then I used this equation:

x_f = x_i + v_i t + 1/2 a (t)^2)

5 = 1/2 (-9.8) t^2

t = 1.20408163



That time is of course wrong...
 

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First of all:
You are using F as a force on the bottom block; but that is clearly false!
The driving force on the bottom block IS THE STATIC FRICTION FROM THE TOP BLOCK ON THE BOTTOM BLOCK!
 
Redo your equation for the bottom block. The applied force "F" acts only on the top block.


(arildno beat me again!)
 
Fixed it. :redface:

Thanks a lot. :)
 

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