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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?

```
4 years ago

Ashutosh
15 Points
```							Balancing torque we get Nx = MgL/2so N = MgL/2xSo friction F = -¥MgL/2xNow 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
```
10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions