Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
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.
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.
The amount of work done by the motor is given by:
[tex]W=\Delta K[/tex]
[tex]W= 1/2*I*\omega f^2-1/2*I*\omega o^2[/tex]
Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:
[tex]\omega f^2=\omega o^2+2*\alpha*\theta[/tex]
But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:
[tex]\tau=I*\alpha[/tex]
[tex]\alpha=\tau/I[/tex] => [tex]\alpha = 0.5rad/s^2[/tex]
Now we can calculate the final velocity:
[tex]\omega f = 8.68rad/s[/tex]
Finally, we calculate the total work:
[tex]W= 1/2*I*\omega f^2 = 1883.56J[/tex]
Since the question asked to "Enter your answer in joules to four significant figures.":
W = 1884J
