A block of mass 1.5 kg kept on a rough horizontal surface is given a horizontal velocity of $10…

A block of mass 1.5 kg kept on a rough horizontal surface is given a horizontal velocity of $10 \mathrm{~ms}^{-1}$. If the block comes to rest after travelling a distance of 12.5 m , the coefficient of kinetic friction between the surface and the block is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
  1. 0.2
  2. 0.4
  3. 0.8
  4. 0.6

Solution


$\begin{aligned} & \mathrm{u}=10 \mathrm{~ms}^{-1}, \mathrm{~m}=1.5 \mathrm{~kg} \\ & \mathrm{a}=\frac{\mathrm{f}}{\mathrm{m}}=\frac{\mu \mathrm{mg}}{\mathrm{m}}=\mathrm{Mg}\end{aligned}$ $\therefore$ Stopping distance, $S=\frac{u^2}{2 a}$ $\begin{aligned} & \Rightarrow 12.5=\frac{(10)^2}{2 \times \mu \times 10} \\ & \therefore \mu=0.4 \end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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