A flight attendant pulls her 74.2 N flight bag
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.
Work = Fk*d = 21.6 * 324 = 6998. J.
c. u = Fk/Fv = 21.6/43.1 = 0.50. = Coefficient of kinetic friction.
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.
To find the answers to these questions, we will need to use some concepts from physics.
a) The work done by a force is given by the formula:
Work = Force * Distance * cos(theta)
where theta is the angle between the force and the direction of motion. In this case, the force exerted by the flight attendant is 37.9 N at an angle of 55.2 degrees above the horizontal. The distance covered is 324 m.
Using the formula, we have:
Work = 37.9 N * 324 m * cos(55.2 degrees)
To find the value, we calculate the cosine of 55.2 degrees (which can be done using a calculator or math software), then multiply it by the other values:
Work ≈ 37.9 N * 324 m * cos(55.2) ≈ 8,438 J
So the work done by the flight attendant on the flight bag is approximately 8,438 J.
b) The work done by the force of friction is equal to the force of friction multiplied by the displacement. Since the flight bag is moving at a constant velocity, the net force acting on it must be zero, thus the force of friction is equal in magnitude and opposite in direction to the force exerted by the flight attendant (37.9 N).
So the work done by the force of friction is:
Work friction = Force of friction * Distance
Work friction = 37.9 N * 324 m
Work friction ≈ 12,276 J
Therefore, the work done by the force of friction on the flight bag is approximately 12,276 J.
c) The coefficient of kinetic friction can be found using the formula:
Coefficient of kinetic friction = (Force of friction) / (Normal force)
The normal force is the force exerted by the floor perpendicular to the surface. Since the flight bag is on a level floor, the normal force is equal and opposite to the weight of the bag.
The weight of the bag can be calculated using the formula:
Weight = mass * gravitational acceleration
Given that the weight of the bag is 74.2 N and the gravitational acceleration is typically around 9.8 m/s^2, we have:
Weight = 74.2 N
So the normal force is also 74.2 N.
Using the equation for the coefficient of kinetic friction:
Coefficient of kinetic friction = 37.9 N / 74.2 N
Coefficient of kinetic friction ≈ 0.51
Therefore, the coefficient of kinetic friction between the flight bag and the floor is approximately 0.51.
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.