A) 6 mg/5, downward direction
B) 6 mg/5, upward direction
C) 6 mg/5, horizontal direction
D) Not possible to have acceleration
Correct Answer: A
Solution :
Let coefficient of friction be u. and 3m block is moving down the incline, then Acceleration \[=\frac{\text{net pulling force}~~~}{\text{total mass}}\] \[a=\frac{3mg\sin {{45}^{o}}-3\mu mg\cos {{45}^{o}}-mg\sin {{45}^{o}}}{4m}\] \[=\frac{(2-3\mu )g}{4\sqrt{3}}\] \[=\frac{(2-1.2)g}{4\sqrt{2}}=\frac{g}{5\sqrt{2}}\] \[(\because \mu =0.4)\] \[\Rightarrow \] \[T=6mg/(5\sqrt{2})\] Force exerted by string/on pulley is \[\sqrt{2}T\]as shown in figure. \[\therefore \,\,\,\,F=\frac{6mg}{5}.\]You need to login to perform this action.
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