Answer:
The coefficient of static friction, μₛ, between the trunk and turntable = 0.32
Explanation:
For this motion of the trunk B,
Initial velocity, v₀ = 0
Tangential Acceleration, a = 0.28 m/s²
Time taken, t = 10s
Using equations of motion,
v = v₀ + at
v = 0 + 0.28 × 10 = 2.8 m/s
Frictional force, Fᵣ = μₛN
μₛ = coefficient of static friction,
N = Normal reaction exerted on the trunk B as a result of its weight = mg
Doing a force balance on the trunk B,
Force keeping the trunk B in circular motion must balance the frictional forces.
Force keeping the trunk B in circular motion, F = mv²/r
Fᵣ = F
μₛN = mv²/r but N = mg
μₛmg = mv²/r
μₛg = v²/r
μₛ = v²/gr
μₛ = 2.8²/(9.8 × 2.5) = 0.32
Hope this helps!!!
