WBJEE · Physics · Laws of Motion
Three blocks are pushed with a force \(\mathrm{F}\) across a frictionless table as shown in figure. Let \(\mathrm{N}_{1}\) be the contact force between the left two blocks and \(\mathrm{N}_{2}\) be the contact force between the right two blocks. Then
- A \(F>N_{1}>N_{2}\)
- B \(F>N_{2}>N_{1}\)
- C \(\quad F>N_{1}=N_{2}\)
- D \(F=N_{1}=N_{2}\)
Answer & Solution
Correct Answer
(A) \(F>N_{1}>N_{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=6 \mathrm{a}, \mathrm{N}_{1}=5 \mathrm{a}, \mathrm{N}_{2}=3 \mathrm{a}\) \(\mathrm{F}>\mathrm{N}_{1}>\mathrm{N}_{2}\)
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