A block is lying at rest inside a bus. The maximum acceleration of the bus such that the block remain…
- $1 \mathrm{~ms}^{-2}$
- $0.5 \mathrm{~ms}^{-2}$
- $2 \mathrm{cms}^{-2}$
- $2 \mathrm{~ms}^{-2}$
Solution

Due to acceleration of bus, block experiences a backward acceleration $a$ Block will move if $m a \geq f$

Where, $f=$ friction force $\begin{aligned} & \Rightarrow \quad m a \geq \mu m g \Rightarrow a \geq \mu g=0.2 \times 10 \\ & \Rightarrow \quad a \geq 2 \mathrm{~ms}^{-2}\end{aligned}$ Hence, maximum acceleration of the bus such that the block remain stationary is $2 \mathrm{~m} / \mathrm{s}^2$.
Asked in: AP EAMCET 2022 (08 Jul Shift 2)