Hence the value of frictional force is given 50 N
let's calculate the force along the plane=mgsin15o = 77.6 N
acceleration down the plane = 77.6−5030=0.92m/s2\frac{77.6-50}{30}=0.92m/s^23077.6−50=0.92m/s2
now applying
v2−u2=2aS32−1.52=2×0.92×SS=3.67mv^2-u^2=2aS\\3^2-1.5^2=2\times0.92\times S\\S=3.67mv2−u2=2aS32−1.52=2×0.92×SS=3.67m
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