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Grade: 12
        
4th question in the attachment.........................................
2 months ago

Answers : (2)

Arun
23763 Points
							

Work done by friction at QR = μmgx

In triangle, sin 30° = 1/2 = 2/PQ

PQ = 4 m

Work done by friction at PQ = μmg × cos 30° × 4 = μmg × √3/2 × 4 = 2√3μmg

Since work done by friction on parts PQ and QR are equal,

μmgx = 2√3μmg

x = 2√3 ≅ 3.5 m

Applying work energy theorem from P to R

mg × sin 30° × 4 = 2√3μmg + μmgx

2 = 4√3μ

μ = 0.29

2 months ago
Vikas TU
10469 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 QR
mgh = (μmgCosQ)PQ + μmg*QR 
h = μCosQ*PQ + μmg*QR 
h = 2 
μ= 0.29 
 
2 months ago
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