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Friction — Physics STD 12 Science — Question

Bihar 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/s2.
✓

Answer

  1.  

R1 - 2g = 0

⇒ R1 = 2 × 10 = 20

2a + 0.2 R1 - 12 = 0

⇒ 2a + 0.2(20) = 12

⇒ 2a = 12 - 4 = 8

⇒ a = 4m/s2

$4\text{a}_1-\mu\text{R}_1=0$

$\Rightarrow4\text{a}_1=\mu\text{R}_1=0.2(20)$

⇒ 4a1 = 4

⇒ a1 = 1m/s2

2kg block has acceleration 4m/s2 & that of 4kg is 1m/s2

  1.  

R1 = 2g = 20

$\text{Ma}-\mu\text{R}_1=0$

⇒ 2a = 0.2(20) = 4

⇒ a = 2m/s2

4a + 0.2 × 2 × 10 - 12 = 0

⇒ 4a + 4 = 12

⇒ 4a = 8

⇒ a = 2m/s2

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