JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 25)
A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms$$-$$2)
_9th_January_Morning_Slot_en_25_1.png)
32 N
18 N
23 N
25 N
Explanation
_9th_January_Morning_Slot_en_25_2.png)
At equilibrium,
N = mgcos45o . . . . . . (1)
3 + mgsin45o = P + $$\mu $$N
$$ \Rightarrow $$ 3 + $${{mg} \over {\sqrt 2 }}$$ = P + 0.6 $$\left( {{{mg} \over {\sqrt 2 }}} \right)$$
$$ \Rightarrow $$ 3 + $${{100} \over {\sqrt 2 }}$$ = P + 0.6 $$ \times $$ $${{100} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ P = 3 + $${{40} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ P = 3 + 20 $${\sqrt 2 }$$
$$ \Rightarrow $$ P = 31.28 $$ \simeq $$ 32 N
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