A 25-newton horizontal force northward and a 35-
1 answer:
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Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>
<em> How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
The amount of work done by the motor is given by:
Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:
But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:
=>
Now we can calculate the final velocity:
Finally, we calculate the total work:
Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J