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Question: A small block of mass m is released from rest from point D and slides down DGF and reaches the point...

A small block of mass m is released from rest from point D and slides down DGF and reaches the point F with speed vF. The coefficient of kinetic friction between block and both the surfaces DG and GF is μ, the velocity vF is

A

2g(y−x)\sqrt { 2 g ( y - x ) }

B

2g(y−μx)\sqrt { 2 g ( y - \mu x ) }

C

2gy\sqrt { 2 g y }

D

2g(y2+x2)\sqrt { 2 g \left( y ^ { 2 } + x ^ { 2 } \right) }

Answer

2g(y−μx)\sqrt { 2 g ( y - \mu x ) }

Explanation

Solution

Here mgy – 12\frac { 1 } { 2 } mvF2 = f2s + f2s2

i.e., 12\frac { 1 } { 2 } mvF2 = mgy - μmg cos βx1 - μmg cos γs2

or 12\frac { 1 } { 2 } vF2 = g (y−μx1 s1 s1−μx2 s2 s2)\left( \mathrm { y } - \frac { \mu \mathrm { x } _ { 1 } \mathrm {~s} _ { 1 } } { \mathrm {~s} _ { 1 } } - \frac { \mu \mathrm { x } _ { 2 } \mathrm {~s} _ { 2 } } { \mathrm {~s} _ { 2 } } \right)

or vF = 2g(y−μx)\sqrt { 2 g ( y - \mu x ) }