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Friction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsFriction5 Marks
Question
A 2kg block is placed over a 4kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12N is applied to:
  1. The upper block.
  2. The lower block. Take $g = 10m/s^2.$
✓

Answer

  1.  

$R_1 - 2g = 0$
$\Rightarrow R_1 = 2 \times 10 = 20$
$2a + 0.2 R_1 - 12 = 0$
$\Rightarrow 2a + 0.2(20) = 12$
$\Rightarrow 2a = 12 - 4 = 8$
$\Rightarrow a = 4m/s^2$
$4\text{a}_1-\mu\text{R}_1=0$
$\Rightarrow4\text{a}_1=\mu\text{R}_1=0.2(20)$
$\Rightarrow 4a_1 = 4$
$\Rightarrow a_1 = 1m/s^2$
$2kg$ block has acceleration $4m/s^2$ & that of 4kg is $1m/s^2$
  1.  

$R_1 = 2g = 20$
$\text{Ma}-\mu\text{R}_1=0$
$\Rightarrow 2a = 0.2(20) = 4$
$\Rightarrow a = 2m/s^2$
$4a + 0.2 \times 2 \times 10 - 12 = 0$
$\Rightarrow 4a + 4 = 12$
$\Rightarrow 4a = 8$
$\Rightarrow a = 2m/s^2$​​​​​​​

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