In the figure,the coefficient of friction bet | vedclass

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(D) Given: Mass of block $A$,$m_{A} = \frac{m}{2}$. Mass of block $B$,$m_{B} = m$.Coefficient of friction between $A$ and $B$,$\mu_{A} = 0.2$. Coeffic

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(D) Given: Mass of block $A$,$m_{A} = \frac{m}{2}$. Mass of block $B$,$m_{B} = m$.Coefficient of friction between $A$ and $B$,$\mu_{A} = 0.2$. Coefficient of friction between $B$ and the floor,$\mu_{B} = 0.1$.For the blocks to move together,the maximum acceleration $a$ is provided by the friction between $A$ and $B$.$f_{max} = \mu_{A} m_{A} g = m_{A} a$$a = \mu_{A} g = 0.2 g$.Now,considering both blocks as a single system,the total mass is $M = m_{A} + m_{B} = \frac{m}{2} + m = \frac{3m}{2}$.The force $F$ must overcome the friction between $B$ and the floor,and provide acceleration $a$ to the system.$F - \mu_{B} (m_{A} + m_{B}) g = (m_{A} + m_{B}) a$$F = (m_{A} + m_{B}) a + \mu_{B} (m_{A} + m_{B}) g$$F = (\frac{3m}{2})(0.2 g) + (0.1)(\frac{3m}{2}) g$$F = 0.3 mg + 0.15 mg = 0.45 mg$.Thus,the maximum force is $0.45 mg$.

Similar Questions

The value of $\theta$ is increased gradually from $\theta = 0$. At $\theta = \tan^{-1}(1/2)$,both blocks just start slipping. Then the value of $\mu_2$ is: $(g = 10 \ m/s^2)$

The acceleration of block $A$ varies with time as shown in the figure. Find the value of the coefficient of kinetic friction between block $A$ and block $B$.

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