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(C) Given:Masses $m_1 = 2 \, kg$ and $m_2 = 4 \, kg$.Applied force $F = 10 \, N$.Since the blocks are connected by a spring and move together,they sha

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$\frac{10}{3}$

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(C) Given:Masses $m_1 = 2 \, kg$ and $m_2 = 4 \, kg$.Applied force $F = 10 \, N$.Since the blocks are connected by a spring and move together,they share the same acceleration $a$.Using Newton's second law for the system:$F = (m_1 + m_2) a$$10 = (2 + 4) a$$10 = 6 a$$a = \frac{10}{6} = \frac{5}{3} \, m/s^2$.Now,consider the $2 \, kg$ block. The only horizontal force acting on it is the spring force $T$.$T = m_1 a$$T = 2 	imes \frac{5}{3} = \frac{10}{3} \, N$.Thus,the spring force between the two blocks is $\frac{10}{3} \, N$.

Two blocks of mass $2 \, kg$ and $4 \, kg$ are accelerated with the same acceleration by a force of $10 \, N$ as shown in the figure on a smooth horizontal surface. Then the spring force between the two blocks will be .......... $N$ (the spring is massless).

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