There are two blocks A and B of masses mA=1kg and mB=3kg kept on the table as shown in figure. The coefficient of friction between A and B as well as between B and the surface of the table is 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: (Take g=10ms- 2 )

Question

There are two blocks A and B of masses mA=1kg and mB=3kg kept on the table as shown in figure. The coefficient of friction between A and B as well as between B and the surface of the table is 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: (Take g=10ms- 2 ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-bgqpqtmvaootdvsf.jpg" /> (A) 12N (B) 8N (C) 16N (D) 40N

Tardigrade Admin · Accepted Answer

For 1kg block, amax=(μ (. 1 .) g/(. 1 .))=2ms- 2 So, Fm a x-fg r o u n d=mt o t a l× amax Fmax-μ (.3+1.)g=(.3+1.)2 Fmax=16N

$12 N$

$8 N$

$16 N$

$40 N$

For $1 k g$ block,
$a_{ma x} = \frac{μ ( 1 ) g}{( 1 )} = 2 m s^{- 2}$
So, $F_{ma x} - f_{g ro u n d} = m_{t o t a l} \times a_{ma x}$
$F_{ma x} - μ (3 + 1) g = (3 + 1) 2$
$F_{ma x} = 16 N$

12N

8N

16N

40N

For 1kg block, amax​=(1)μ(1)g​=2ms−2 So, Fmax​−fground​=mtotal​×amax​ Fmax​−μ(3+1)g=(3+1)2 Fmax​=16N

$12 N$

$8 N$

$16 N$

$40 N$

For $1 k g$ block,
$a_{ma x} = \frac{μ ( 1 ) g}{( 1 )} = 2 m s^{- 2}$
So, $F_{ma x} - f_{g ro u n d} = m_{t o t a l} \times a_{ma x}$
$F_{ma x} - μ (3 + 1) g = (3 + 1) 2$
$F_{ma x} = 16 N$