Q

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)

$ms^{-2}$

Views

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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