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Two masses m(1) =10 Kg and m(2)=5kg are ...

Two masses `m_(1) =10 Kg` and `m_(2)=5kg` are connected by an ideal string as shown in the figure. The coefficient of friction between `m_(1)` and the surface is `mu=0.2` Assuming that the system is released from rest calculate the velocity of blocks when `m_(2) has descended by `4m`. `(g=10 m//s^(2))`
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Text Solution

Verified by Experts

The correct Answer is:
D
`v_(m_1) =v_(m_(2))`
`d_(m_1) =h_(m_2) =4m`
Using the equation,
`E_(i)-E_(f) =`
Work done against friction
`0-[(1)/(2) xx (10 + 5) (v^(2))-5 xx 10 xx 4]`
`=0.2 xx 10 xx 10 xx 4`
Solving we get,
`v=4m//s`
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Knowledge Check

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