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A talented PHY210 student is tuning her car. She increase the engine speed from 660 rpm very (69.1 rad/s) to 4500 rpm (471 rad/s). If the flywheel has a moment of inertia of 0.525 kg-m2 calculate the work done by the engine on the flywheel during this change in angular velocity. Select one: o a. 105 b, 1.43 x 104 J c. 1.44 x 103 O d. 5.70 x 104 J e. 1.14 x 105J

Respuesta :

Answer:

d. 5.7 x 10⁴ J

Explanation:

I = moment of inertia of the flywheel = 0.525 kg-m²

w₀ = initial angular speed of the flywheel = 69.1 rad/s

w = final angular speed of the flywheel = 471 rad/s

W = work done by the engine on the flywheel

Work done by the engine is given as

W = (0.5) I (w² - w₀²)

Inserting the values

W = (0.5) (0.525) (471² - 69.1²)

W = 5.7 x 10⁴ J

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