The coefficient of static friction between the block of 2 kg and the table shown in figure is `mu_s=02.` What should be the maximum value of m so that the blocks does not move? Take `g=10 m//s^2.` The string and the pulley are light and smooth.

Options

A. 0.4 kg
B. 0.2 kg
C. 0.1 kg
D. 0.5 kg

Answer

A

Explanation

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`