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

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Aicharya · Accepted Answer

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 force, hence $10 \mathrm{~kg}$ block will start motion due to application of $100 \mathrm{~N}$ force. Due to motion, kinetic friction force $\begin{aligned} f_k & =\mu_k R=0.4 \mathrm{mg} \ & =0.4 \times 10 \times 9.8=39.2 \mathrm{~N} \end{aligned}$ $40 \mathrm{~kg}$ body experiences a force of $f_k=39.2 \mathrm{~N}$ $\therefore$ Acceleration of $40 \mathrm{~kg}$ slab $a=\frac{f_k}{40}=\frac{39.2}{40}=0.98 \mathrm{~ms}^{-2}$

If \(100 \mathrm{~N}\) force is applied to \(10 \mathrm{~kg}\) block as shown in the diagram, the…

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