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A block of mass m is kept on a smooth surface. It is also in contact of a thin plank of mass M = 2m, one end of which is hinge to a wall as shown in the figure. The length of the plank is L. The block is given an initial velocity v0, towards right and initially the block is at a distance x0 from hinge. Coefficient of friction between the plank and block is . Assume that throughout the motion block remains below the rod, then at what position from the fixed point it will stop?

A block of mass m is kept on a smooth surface. It is also in contact of a thin plank of mass M = 2m, one end of which is hinge to a wall as shown in the figure. The length of the plank is L. The block is given an initial velocity v0, towards right and initially the block is at a distance x0 from hinge. Coefficient of friction between the plank and block is . Assume that throughout the motion block remains below the rod, then at what position from the fixed point it will stop?

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Grade:12

1 Answers

Ashutosh
16 Points
4 years ago
Balancing torque we get Nx = MgL/2
so N = MgL/2x
So friction F = -¥MgL/2x
Now acceleration a = -¥MgL/2xm = VdV/dx
 
Solve by integration with limits of x be 0 to x and limit of v be v0 to 0

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