A 18.5 kg door that is 1.02 m wide opens slowly as a pull of constant magnitude 53.3 N is exerted on the doorknob. This applied force, which is always perpendicular to the door, is necessary to balance the frictional torque in the hinges. How large must an applied force of constant magnitude be in order to open the door through an angle of 90.0° in 0.474 s?

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A 18.5 kg door that is 1.02 m wide opens slowly as a pull of constant magnitude 53.3 N is exerted on the doorknob. This applied force, which is always perpendicular to the door, is necessary to balance the frictional torque in the hinges. How large must an applied force of constant magnitude be in order to open the door through an angle of 90.0° in 0.474 s?

 

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