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<span>Taking into account the information above, we know the average mass of the bucket of water may be m=20-5/2=17.5kg. As the bucket of water is pulled at a "constant velocity" the work required to raise the bucket to the platform transformed into the potential energy of the bucket of water. That is why it should be W=mgh=17.5*9.8*40=6860J</span>
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
= r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
= r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as


= 0.42 m/s²