A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se conds after starting from rest? 312 J
237 J
156 J
106 J
1 answer:
First you do 2.5 n-m / 2 kg-m2 This equals 1.25 rad/s^2 Then multiply rad/s^2 by 10 because of the amount of seconds This equals 12.5 rad/s^2 Then do this following equation : (1/2)Iwf^2 Do (0.5) 12.5 * 12.5 * 2 This equals 156.25 J (meaning that the answer is C: 156 J)
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