A horizontal force just sufficient to move a body of mass $4 \mathrm{~kg}$ lying on a rough horizontal…
- $6 ms^{-2}$
- $8 ms^{-2}$
- $2 ms^{-2}$
- $4 ms^{-2}$
Solution

$\begin{aligned} & F=f_s=\mu_s N \\ & =0.4 \times 4 \times 10=32 \mathrm{~N}\end{aligned}$ Finally,

$\begin{aligned} & \mathrm{F}-\mathrm{f}_{\mathrm{k}}=\mathrm{ma} \\ & 32-\mu_{\mathrm{k}} \mathrm{mg}=\mathrm{ma} \\ & \Rightarrow \quad 32-0.6 \times 40=4 \mathrm{a} \\ & \Rightarrow \quad 8=4 \mathrm{a} \\ & \Rightarrow \mathrm{a}=2 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
Asked in: AP EAMCET 2015