JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 25)
A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is __________ N. (Round off to the Nearest Integer) [Take g = 10 ms-2 ]
_17th_March_Evening_Shift_en_25_1.png)
Vastaus
30
Selitys
_17th_March_Evening_Shift_en_25_2.png)
$$ \because $$ f = T
$$ \Rightarrow $$ $$\mu$$N = T
$$ \Rightarrow $$ $$\mu$$(90 $$-$$ T) = T
$$ \Rightarrow $$ 0.5 (90 $$-$$ T) = T
$$ \Rightarrow $$ 90 $$-$$ T = 2T
$$ \Rightarrow $$ 3T = 90
$$ \Rightarrow $$ T = 30 N