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How to solve this problem- - Laws of motion - JEE Main-3

A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F=20N, making an angle of 30^{\circ} with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is \mu =0.2. The difference between the accelerations of the block, in case (B) and case (A) will be : (g=10ms^{-2})

 

  • Option 1)

    0.4 ms^{-2}

  • Option 2)

    3.2 ms^{-2}

  • Option 3)

    0.8 ms^{-2}

  • Option 4)

    ms^{-2}

 
Answers (1)
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Newton's 2nd Law -

F\propto \frac{dp}{dt}

F=\tfrac{kdp}{dt} 

F=\tfrac{d\left (mv \right )}{dt} 

F=\tfrac{m\left (dv \right )}{dt}

\frac{dv}{dt}=a

Therefore  F=ma

- wherein

K=1 in C.G.S & S.I

Force can be defined as rate of change of momentum.

 

 

                                                                        

N_{1}=50+10=60N                                    N_{2}=50-10=40N

F_{lim}=0.2\times 60 =12 N                                F_{lim}=0.2\times 40=8N

F_{applied}>12N                                                 F_{applied}>8N

f_{1}=12 N                                                            f_{1}=8N

a_{2}=\frac{17.328}{5}=\frac{9.32}{5}

a_{1}=\frac{10\sqrt{3}-12}{5}=\frac{17.32-12}{5}=\frac{5.32}{5}

a_{1}-a_{2}=\frac{5.32}{5}-\frac{9.32}{5}=\frac{-4}{5}

|a_{1}-a_{2}|=\frac{4}{5}

 


Option 1)

0.4 ms^{-2}

Option 2)

3.2 ms^{-2}

Option 3)

0.8 ms^{-2}

Option 4)

ms^{-2}

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