Two blocksA and B of masses m_{A}=1Kgand  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)
V Vakul

\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

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