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Two blocksA and B of masses $m_{A}=1Kg$ and $m_{B}=3Kg$ re kept on the table as shown in figure . The coefficient of friction between A and B is 0,2 .and between B and the surface of the table is also 0.2 The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : $\left [ Take\, \, g=10m/s^{2} \right ]$

Option 1)
$8N$
Option 2)
$16N$
Option 3)
$40N$
Option 4)
$12N$

Answers (1)

$\mu_{ AB}=0.2$

$\mu_ {B surface}=0.2$

$M_A=1Kg$

$M_B=3Kg$

for 1 Kg block

$a\, \, _{max}=\frac{\mu \left ( 1 \right )\left ( g\right )}{1}=\left ( .2 \right )\left ( 1 \right )\left ( 10 \right )=2m/s^{2}$

Now,

$f=\left ( f_{r} \right )\, ground=m_{t}\, a\, \,_ {max}$

$f=\left ( 4 \right )\left ( 2 \right )+\mu \left ( 4 \right )\left ( g \right )$

$f=8+\left ( .2 \right )\left ( 4 \right )\left ( 10 \right )$

$f=8+8$

$f=16N$

Option 1)

$8N$

Option 2)
$16N$
Option 3)

$40N$

Option 4)

$12N$

Posted by

Vakul

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