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`        4th question in the attachment.........................................`
one month ago

Arun
23520 Points
```							Work done by friction at QR = μmgxIn triangle, sin 30° = 1/2 = 2/PQPQ = 4 mWork done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmgSince work done by friction on parts PQ and QR are equal,μmgx = 2√3μmgx = 2√3 ≅ 3.5 mApplying work energy theorem from P to Rmg × sin 30° × 4 = 2√3μmg + μmgx2 = 4√3μμ = 0.29
```
one month ago
Vikas TU
10070 Points
```							As energy lost over PQ = energy lost over QR. μmgCosQPQ =  μmg QR SinQ = 1/2 PQ = 4m QR = 4cos30 = 3.5m Decrease in PE = loss of energy due to friction in PQ and QRmgh = (μmgCosQ)PQ + μmg*QR h = μCosQ*PQ + μmg*QR h = 2 μ= 0.29
```
one month 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