The coefficient of static friction, mu(s...

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

4.0kg.

0.2kg.

0.4kg.

2.0kg.

Text Solution

Verified by Experts

The correct Answer is:

`T=M_(B)g=mu_(s)M_(A)g`
`therefore M_(B)=mu_(s)M_(A)=0.2xx2=0.4kg`

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