Complete question :
A flight attendant pulls her 70 N flight bag a distance of 318 m along a level airport floor at a constant velocity. The force she exerts is 41 N at an angle of 57° above the horizontal.
(a) Find the work she does on the flight bag.
(b) Find the work done by the force of friction on the flight bag.
(c) Find the coefficient of kinetic friction between the flight bag and the floor.
Answer:
7107 J ; - 7107 J ; 0.55
Step-by-step explanation:
Given that :
Distance, d = 318m
Applied force = 41 N
θ = 57°
A.) Workdone = Force exerted along direction of motion
Workdone = applied Force * distance * cosθ
Workdone = 41 * 318 * cos57 = 7101.0037
Workdone = 7,101 J
B.) Workdone by force of friction on flight bag:
- 7,101 J (since the body moves at constant velocity)
C.)
Coefficient of kinetic friction (μ) = Frictional force / normal reaction)
μ = F / N
Frictional force, F = Workdone by friction / distance
F = 6200 / 318
F = 19.47N ;
Ff = weight of Flight bag = 70 N
N = Ff - Fsinθ
N = 70 - applied Force sinθ
N = 70 - 41sin57
N = 70 - 34.385493
N = 35.614506
μ = 19.47 / 35.614506
μ = 0.5466873