a flight attendant pulls her 70.0 N flight bag a distance of 253m along a level airport floor at constant velocity. The force she exerts is 40.0 N at the angle of 52.0 degree above the horizontal. what is the work she does on the flight bag and the work done by the force of friction on the flight bag.
help please..............
2007-11-17 14:27:44 · 3 answers · asked by 832 1 in Science & Mathematics ➔ Physics
Use Newton's Second Law in each dimension.
Let P be the pulling force of 70.0 N.
Vertical:
Since the bag has no vertical acceleration, the net vertical force on the bag is zero. The three forces on the bag are mg down, the normal force applied by the floor up, and P sin 52 degrees up.
0 = N + P sin 52 - mg
Solve for N.
Now horizontally: again, the bag has no horizontal acceleration so there's no net force. The two forces on the bag are the retarding friction force uN (where u is the coefficient of kinetic friction between the bag and the floor) and the other component of the pulling force, P cos 52.
0 = P cos 52 - uN
Solve for u, if you want too, but you don't actually need it.
Work is the scalar product of force and displacement; in this case, simply F d. The work done by her and the work done by friction are equal and opposite.
W = P cos 52 d = u N d
HTH!
2007-11-17 14:37:54 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
This is a simple exercise in vectors. The force she/he exerts is 40N times the cosine of 52 degrees, since (s)he doesn't lift the bag off the ground. The work done is force times distance for constant force and velocity. The work done by friction is the same as the work done by the attendant.
2007-11-17 14:39:01 · answer #2 · answered by banjoman 6 · 0⤊ 0⤋
frictional force =40 N
2007-11-17 14:39:27 · answer #3 · answered by nitish 1 · 0⤊ 1⤋