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The coefficient of static friction betw...

The coefficient of static friction between the block of 2 kg and the table shown in figure is `mu_s=0.2.` What should be the maximum value of m so tht the blocks do not move? Take `g=10 m/s^2.` The string and the pulley are light and smooth.

A

0.4 kg

B

0.2 kg

C

0.1 kg

D

0.5 kg

Text Solution

Verified by Experts

The correct Answer is:
A
Consider the equilibrium of the block of mass m. The forces on this block are
a. mg downward by the earth and
T upward by the string.
hence `T-mg=0 or, T=mg`…….i
Now consider the equilibrium of the 2 kg block. The forces on this block are
a. T towards rightby the string,
b. f towards left (friction)by the table,
c. 20 N downward (weight) by the earth and
d. N upward (normal force) by the table
For vertical equilibrium of this block,
N=20N...........ii
A m is the largest mass which can be used without moving the system, the friction is limiting
Thus, `f=mu_sN` ...........iii
For horizontal equilibrium of the 2 kg block
f=T
Using equations i, iii and iv
`mu_sN=mg`
or `02xx20N=mg`
or, `m=(02xx20)/10kg=04kg`
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