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```
a man of mass 50kg standing on one end of a stationary wooden plank resting on a frictionless surface the mass of the plank is 100kg and its length is 75m and the coefficient of friction between the man and plank is 0.2 find the least possible time in which the man reach the other end starting from rest and stopping at the other end

```
9 months ago

Arun
25768 Points
```							Dear student a = 0.2 *10 + (50*0.2*10/10) = 2 + 10 = 12 Vmax = √(2*12*150) = 60 Hence 60 *(t/2) = 300t/2 = 5t = 10 sec Hope it helps
```
9 months ago
Vikas TU
14146 Points
```							Dear student a1=μmg/m=2m/s^2a2=(μmg)M=1m/s^2Plank as frame of reference Plank comes to resttanα=3=h/th=3tfor minimum time, man will acceleration and retard with a1/2 for equal time interval t Area of v vs t graph = displacement1/2(2t)(3t)=75t=5minimum time =2t=10sec
```
9 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