AP EAMCET · PHYSICS · Laws of Motion
If \(100 \mathrm{~N}\) force is applied to \(10 \mathrm{~kg}\) block as shown in the diagram, the acceleration of \(40 \mathrm{~kg}\) slab is
- A \(1.65 \mathrm{~ms}^{-2}\)
- B \(0.98 \mathrm{~ms}^{-2}\)
- C \(0.5 \mathrm{~ms}^{-2}\)
- D \(0.25 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(B) \(0.98 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Static friction force between \(10 \mathrm{~kg}\) and \(40 \mathrm{~kg}\) block, \(\begin{aligned} F_s & =\mu_s R=0.6 \times m g \\ & =0.6 \times 10 \times 9.8=58.8 \mathrm{~N} \end{aligned}\) Here, we see that the applied force \((F=100 \mathrm{~N})\) is greater than friction…
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