Work done by friction at QR=µmgx In triangle, sin‌30°=
1
2
=
2
PQ
⇒PQ=4m Work done by friction at PQ=µmg×Cos‌30°×4 =µmg×
√3
2
×4=2√3µmg Since work done by friction on parts PQ and QR are equal, µmgx=2√3umg ⇒x=2√3≅3.5m Using work energytheorem mg‌sin‌30°×4=2√3µmg+µmgx ⇒2=4√3µ ⇒µ=0.29