The coefficient of static friction, μ s , between block A of mass 2kg and the table as shown in the figure is 0.2 . The maximum mass value of block B so that the two blocks do not move is (The string and the pulley are assumed to be smooth and massless g=10 ms- 2)

Question

The coefficient of static friction, μ s , between block A of mass 2kg and the table as shown in the figure is 0.2 . The maximum mass value of block B so that the two blocks do not move is (The string and the pulley are assumed to be smooth and massless g=10 ms- 2) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xqiffotib42dadvf.jpg" /> (A) 0.4 kg (B) 2.0 kg (C) 4.0 kg (D) 0.2 kg

Tardigrade Admin · Accepted Answer

Condition for not moving the block is that the tension in the string should be equal to static frictional force, here tension will be equal to mBg mB g=μ s mA g ⇒ mB=μ s mA or, mB=0.2× 2=0.4 kg

Q. The coefficient of static friction, $μ_{s},$ between block $A$ of mass $2 k g$ and the table as shown in the figure is $0.2$ . The maximum mass value of block $B$ so that the two blocks do not move is (The string and the pulley are assumed to be smooth and massless $g = 10 m s^{- 2})$

$0.4 k g$

$2.0 k g$

$4.0 k g$

$0.2 k g$

Condition for not moving the block is that the tension in the string should be equal to static frictional force, here tension will be equal to $m_{B} g$
$m_{B} g = μ_{s} m_{A} g \Rightarrow m_{B} = μ_{s} m_{A} or, m_{B} = 0.2 \times 2 = 0.4 k g$

Q. The coefficient of static friction, μs​, between block A of mass 2kg and the table as shown in the figure is 0.2 . The maximum mass value of block B so that the two blocks do not move is (The string and the pulley are assumed to be smooth and massless g=10ms−2)

0.4kg

2.0kg

4.0kg

0.2kg