A flight attendant pulls her 74.2 N flight bag

Question

a distance of 324 m along a level airport floor
at a constant velocity. The force she exerts is
37.9 N at an angle of 55.2� above the horizon-
tal.
a) Find the work she does on the flight bag.
Answer in units of J
024 (part 2 of 3) 10.0 points
b) Find the work done by the force of friction
on the flight bag.
Answer in units of J
025 (part 3 of 3) 10.0 points
c) Find the coefficient of kinetic friction be-
tween the flight bag and the floor.

Answer 1

a. Work = 37.9*cos55.2 * 324 = 7,008 J.

b. Wb = 74.2 N. = Wt. of bag.

Fb = 74.2N @ 0o. = Force of bag.
Fp=74.2*sin(0)=0=Force parallel to floor
Fv = 74.2*cos(0)-37.9*sin55.2 = 43.1 N =
Force perpendicular to floor.

Fap*cos55.2 - Fp - Fk = ma = m*0 = 0.
37.9*cos55.2-0-Fk = 0
Fk = 37.9*cos55.2 = 21.6 N. = Force of
kinetic friction.

Answer 2

Work = Fk*d = 21.6 * 324 = 6998. J.

c. u = Fk/Fv = 21.6/43.1 = 0.50. = Coefficient of kinetic friction.

Answer 3

gugjggggggggggggggg\ggga. Work = 37.9*cos55.2 * 324 = 7,008 J.

b. Wb = 74.2 N. = Wt. of bag.

Fb = 74.2N @ 0o. = Force of bag.
Fp=74.2*sin(0)=0=Force parallel to floor
Fv = 74.2*cos(0)-37.9*sin55.2 = 43.1 N =
Force perpendicular to floor.

Fap*cos55.2 - Fp - Fk = ma = m*0 = 0.
37.9*cos55.2-0-Fk = 0
Fk = 37.9*cos55.2 = 21.6 N. = Force of
kinetic friction. bhghWork = Fk*d = 21.6 * 324 = 6998. J.

c. u = Fk/Fv = 21.6/43.1 = 0.50. = Coefficient of kinetic friction.