Consider the system shown below. A horizontal | vedclass

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(A) Let the acceleration of the system of blocks $X$ and $Y$ be $a$.Since the blocks move together,the total mass is $M = m_X + m_Y = 8 \,kg + 2 \,kg

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) Let the acceleration of the system of blocks $X$ and $Y$ be $a$.Since the blocks move together,the total mass is $M = m_X + m_Y = 8 \,kg + 2 \,kg = 10 \,kg$.The acceleration of the system is given by $a = \frac{F}{M} = \frac{F}{10} \,ms^{-2}$.For block $Y$ to not slip downwards,the upward frictional force $f$ must balance its weight $m_Y g$.$f = m_Y g = 2 	imes 10 = 20 \,N$.The frictional force is also given by $f = \mu R$,where $R$ is the normal reaction between blocks $X$ and $Y$.From the horizontal motion of block $Y$,the normal reaction $R$ provides the necessary acceleration $a$ to block $Y$:$R = m_Y a = 2 	imes \frac{F}{10} = \frac{F}{5}$.Substituting the values into the friction equation:$f = \mu R \Rightarrow 20 = 0.5 	imes \frac{F}{5}$.$20 = \frac{F}{10} \Rightarrow F = 200 \,N$.Thus,the minimum value of $F$ is $200 \,N$.

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