4th question in the attachment.........................................
4th question in the attachment.........................................
sri harshitha honey , 6 Years ago
Grade 12
Mechanics> 4th question in the attachment..............
2 Answers
Arun
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
Vikas TU
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
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
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