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Question:

kela000
kela000
Student
Community / Jr. College

Attendant pulls her 74.2N flight bag a distance of 324m along a floor at constant velocity.Force exerts is 37.9N at angle 55.2degrees above horizontal

A.) find the work she does on the flight bag. Answer in units of J.

B.) Find the work done by the force of friction on the flight bag. Answer in units of J.

C.) Find the coefficient of kinetic friction between the flight bad and the floor.

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Posted by kela000 on Monday November 9, 2009 at 8:47 AM and tagged with energies, math.

Answers:

  1. neela
    neela Teacher
    Graduate School

    eNotes Editor

    a.The work the flight attendant does on the flight bag =  component of force she pulls the bag along the floor * the displacement of the bag = (37.9N cos 55.2 degree)*(324 meter) = 7008.1343 J

    b. The force of friction is equal but opposite to the component of the exerted force resolved along the floor in the direction of displacement * displacement =  (- 37*9N cos 55.2)*(324) cos (angle between frictional force and direction of displacement)= (-21.63004N)(324)*cos(180)=7008.1343 J

    c.The frictional force, F is given by: F1= -The coefficient offriction* Normal force. Normal force = mg-37.9.sin 55.2 deg

    F1= - M(74.2-37.9sin 55.2) = 43.07844*M, where M is the coefficient of  kinetic friction.

    The pulling force is F2 =  component of 37.9N along the direction of displacement of the bag.

    Since there is no acceleration, the net force alon g the floor is zero. Or F1+F2=0 or

    -43.07844*M+37.9N cos 55.2 = 0

    M= 37.9 cos 44.2/43.07844 =0.5021

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    Posted by neela on Monday November 9, 2009 at 9:55 AM