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a hanging block of mass m is connected to an identical block of mass m …

Question

a hanging block of mass m is connected to an identical block of mass m on a horizontal surface with negligible friction by a light cord over an ideal pulley, as shown. when the blocks are released, the magnitude of the acceleration of each block is a. if the block on the surface is replaced with a block of mass 5m, what will be the magnitude of the acceleration of each block after the blocks are released?
a $\frac{1}{6}a$
b $\frac{1}{5}a$
c $\frac{1}{3}a$
d $\frac{5}{2}a$

Explanation:

Step1: Analyze initial - case

For the initial case with two blocks of mass $M$, using Newton's second law $F = ma$. The net - force acting on the system is the weight of the hanging block, $F = Mg$, and the total mass of the system is $M + M=2M$. So, $Mg=(M + M)a$, which simplifies to $a=\frac{g}{2}$.

Step2: Analyze second - case

When the block on the surface has mass $5M$, the net - force acting on the system is still the weight of the hanging block $Mg$, and the total mass of the system is now $M + 5M = 6M$. Let the new acceleration be $a'$. Using Newton's second law $F = ma$, we have $Mg=(M + 5M)a'$. Solving for $a'$, we get $a'=\frac{Mg}{6M}=\frac{g}{6}$.
Since $a=\frac{g}{2}$, then $a'=\frac{1}{3}a$.

Answer:

C. $\frac{1}{3}a$