First slide

Static friction

Question

In the diagram shown in figure, match the following columns (Take, g = 10 ms^-2)
Column I
(A) Normal reaction
(8) Force of friction
(C) Acceleration of block
Column II
(p) 12 SI unit
(q) 20 SI unit
(r ) zero
(s) 2 SI unit

Moderate

A

A B C
p q r
B

A B C
q p r
C

A B C
q p s
D

A B C
p s r

Solution

$N = m g - 20 \sqrt{2} \sin 45 ° = 20 N, f = μ_{k} N = 12 N$
Since, $20 \sqrt{2} \cos 45 ° > μ N$ block will move and its acceleration will be
$a = \frac{20 \sqrt{2} \cos 45 ° - μ_{k} N}{m} = \frac{20 - 12}{4} = 2 {ms}^{- 2}$
Hence, $A \to q, B \to p, C \to s$ .

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