When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting
1 answer:
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Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².