A flight attendant pulls her 75.0 N flight bag a distance of 241.0 m along a level airport floor at constant speed. The force she exerts is 438 N at an angle of 66.0° above the horizontal. Find:
The work she does on the flight bag.
The work done by the force of friction on the flight bag.
The coefficient of kinetic friction between the flight bag and the floor.
How would you solve this problem? With what formulas?
2006-10-31 08:15:58 · 4 answers · asked by gunder53534 2 in Science & Mathematics ➔ Physics
First draw a force diagram showing the floor, bag and the force from the F/A. The normal component of the force of the bag on the floor * the Cf will equal the horizontal force of the floor on the bag. The Fh times the distance will equal the work
This will equal the friction work.
There may be a problem with your input data. Immediately I see that the vertical component of the applied force (438sin66) =400N is more than the weight of the bag, so it will immediately leave the floor, eliminating friction and energy losses, etc..........!
2006-10-31 08:28:54 · answer #1 · answered by Steve 7 · 0⤊ 0⤋
you can use work done = F x d.
since the force she applied is at an angle, you must realise that there is a horizontal component and a vertical component of tt diagonal force that keeps the flight bag moving forward and held up. So, to find the horizontal component, it is 438 x cos 66 (take the Force Diagram to be a right angled triangle. Hypotenus = 438, horizontal force is the adj line.)
The work done on the flight bag = work done by friction since it travels at constant speed.
I THINK the above solves the 1st 2 qns.
2006-10-31 08:25:03 · answer #2 · answered by luv_phy 3 · 0⤊ 0⤋
Work is force over distance F.d
Force is only the part in direction of travel (need a little trigonometry)
Bag travels at constant speed so no work is done there
2006-10-31 08:22:00 · answer #3 · answered by bubsir 4 · 0⤊ 0⤋
Fk = zero.two m = five kg d = 10 m g = nine.eight m/s squared (or in my case, my trainer has us circular to ten m/s squared Ff = Fk x m x g Ff = zero.two x five kg x 10 m/s squared = 10 W = F x d W = 10 (reply from above) x 10 m = one hundred J however considering that the frictional drive acts in opposition to the course the sled is journeying, the reply is -one hundred J
2016-09-01 05:14:25 · answer #4 · answered by kaufmann 4 · 0⤊ 0⤋