Work on sliding bag on aircraft

AI Thread Summary
The discussion focuses on calculating the work done by a flight attendant pulling her bag and the work done by friction. The flight attendant exerts a force of 32.0 N at an angle of 52° over a distance of 258 m. The correct formula for calculating the work done by the attendant is 32 N * cos(52°) * 258 m, leading to a result of approximately 1345.65 J. Since the bag moves at a constant speed, the work done by friction is equal in magnitude but negative, indicating that it cancels out the work done by the attendant. Overall, the total external work done on the bag is zero, as there is no change in kinetic energy.
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Can anyone help me with this problem?

A flight attendant pulls her 70.0 N flight bag a distance of 258 m along a level airport floor at a constant speed. The force she exerts is 32.0 N at an angle of 52.0° above the horizontal. Find the work she does on the flight bag and the work done by the force of friction of the flight bag.

I was able to make a free body diagram and see that the weight of the flight bag is 70 Newtons, the displacement is 258 m. But when I solved for W, which is force x displacement x cosine 52 degrees I got the wrong answer.
(70 x 258 x cos52).
 
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your answer looks right to me, why do you say that it is wrong.

1345.65J
 
Isn't the calculation 32*cos52*258? Since the speed was constant, friction did the same amount of work.
 
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Because the assignment is online, I'm able to put it into see if I get it right or not.
 
The 70 N weight has nothing to do withthe flight attendant's Work;
most of it is canceled by the Surface compression Force by the Floor (F_N).
vertigo is right, there.
But Friction did a NEGATIVE amount of Work
(total external Work done on the bag = change of its KE = 0).
 
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