A block of mass $10 \mathrm{~kg}$ placed on the rough horizontal surface having coefficient of friction…
- $10 \mathrm{~ms}^{-1}$
- $5 \mathrm{~ms}^{-2}$
- $15 \mathrm{~ms}^{-2}$
- $0.5 \mathrm{~ms}^{-2}$
Solution
Here $f_{\max }=\mu N=\mu m g$ $=0.5 \times 10 \times 10=50 \mathrm{~N}$ $\Rightarrow a=\frac{\text { net force }}{\text { mass }}$ $=\frac{100-50}{10}=5 \mathrm{~m} / \mathrm{sec}^2$Asked in: NEET 2002