In the figure,the coefficient of friction bet | vedclass

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Question

(D) For both blocks to move together,block $A$ must move with the same acceleration $a$ as block $B$. The only force causing acceleration on block $A$

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(D) For both blocks to move together,block $A$ must move with the same acceleration $a$ as block $B$. The only force causing acceleration on block $A$ is the static friction force $f_s$ from block $B$.The maximum static friction force between $A$ and $B$ is $f_{s,max} = \mu_2 m_A g = 0.3 	imes 20 	imes 10 = 60 	ext{ N}$.The maximum acceleration $a_{max}$ of block $A$ is $a_{max} = \frac{f_{s,max}}{m_A} = \frac{60}{20} = 3 	ext{ m/s}^2$.Now,consider the system of both blocks $(A+B)$ moving together with acceleration $a_{max} = 3 	ext{ m/s}^2$. The external force $F$ must overcome the kinetic friction $f_k$ between block $B$ and the floor.The normal force on the floor is $N = (m_A + m_B)g = (20 + 40) 	imes 10 = 600 	ext{ N}$.The kinetic friction force is $f_k = \mu_1 N = 0.2 	imes 600 = 120 	ext{ N}$.Applying Newton's second law to the system $(A+B)$:$F - f_k = (m_A + m_B) a_{max}$$F - 120 = (20 + 40) 	imes 3$$F - 120 = 60 	imes 3$$F - 120 = 180$$F = 300 	ext{ N}$.

Similar Questions

Find the maximum force $F$ (in $N$) for which both the blocks will move with the same acceleration.

The coefficient of friction between two blocks is $0.2$,and the coefficient of friction between block $B$ and the surface is $0.5$. If a force of $25 \ N$ is applied to block $B$ (mass $A = 2 \ kg$,mass $B = 8 \ kg$),what is the magnitude of the frictional force between the two blocks in $N$?

Imagine a situation in which the horizontal surface of block $M_0$ is smooth and its vertical surface is rough with a coefficient of friction $\mu$. Identify the correct statement$(s)$.

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