MHT CET · Physics · Laws of Motion
Two blocks of masses 6 kg and 4 kg are placed in contact with each other on a smooth surface as shown. If a force of 5 N is applied on a heavier block, the force on the lighter block is
- A 5 N
- B 4 N
- C 2 N
- D 1 N
Answer & Solution
Correct Answer
(C) 2 N
Step-by-step Solution
Detailed explanation
Given: \(\mathrm{m}_1=6 \mathrm{~kg}, \mathrm{~m}_2=4 \mathrm{~kg}\) and \(\mathrm{F}=5 \mathrm{~N}\)
\(\begin{array}{ll}
& F=\left(m_1+m_2\right) a \\
\therefore \quad & a=\frac{F}{\left(m_1+m_2\right)}=\frac{5}{(6+4)}=0.5 \mathrm{~m} / \mathrm{s}^2
\end{array}\)
\(\begin{aligned}
\therefore \quad \text { Force on second mass } & =\mathrm{m}_2 \times \mathrm{a} \\
& =4 \times 0.5 . \\
& =2 \mathrm{~N}
\end{aligned}\)
\(\begin{array}{ll}
& F=\left(m_1+m_2\right) a \\
\therefore \quad & a=\frac{F}{\left(m_1+m_2\right)}=\frac{5}{(6+4)}=0.5 \mathrm{~m} / \mathrm{s}^2
\end{array}\)
\(\begin{aligned}
\therefore \quad \text { Force on second mass } & =\mathrm{m}_2 \times \mathrm{a} \\
& =4 \times 0.5 . \\
& =2 \mathrm{~N}
\end{aligned}\)
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