contestada

A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following: a) the work she does on the flight bag b) the work done on the flight bag c) the coefficient of kinetic friction between the flight bag and the floor.

Respuesta :

boffeemadrid

Answer:

6230.49413 J

0 J

0.63998

Explanation:

F = Force = 40 N

[tex]\theta[/tex] = Angle = 52°

Work done is given by

[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]

The work she does on the flight bag is 6230.49413 J

The work done on the flight bag will be the opposite of the work done by the flight attendant

[tex]W=-6230.49413\ J[/tex]

So net work will be

[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]

The net work done on the flight bag is 0 J

Coefficient of friction is given by

[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]

The coefficient of friction is 0.63998

snehashish65

(a)  The work done by the attendant on the flight bag is 6230.49 J.

(b)  The work on the flight bag is 0 J .

(c)  The coefficient of kinetic friction between the flight bag and the floor is 0.35.

Given data:

The weight of flight bag is, W = 70.0 N.

The distance covered by the bag is, d = 253 m.

The magnitude of force exerted on bag is, F = 40.0 N.

The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].

Work done defined as the product of force and distance covered due to applied force.

(a)

The work done on the flight bag is given as,

[tex]W'=F \times dcos\theta[/tex]

Solving as,

[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]

Thus, the work done by the attendant on the flight bag is 6230.49 J.

(b)

The work done on the flight bag will be the opposite of the work done by the flight attendant. So,

W'' = - W

W'' = - 6230.49 J

Then net work done on the flight bag is,

[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]

Thus, the net work on the flight bag is 0 J .

(c)

The expression for the frictional force is given as,

[tex]f = \mu \times W[/tex]

[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,

[tex]f=Fcos\theta[/tex]

So,

[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]

Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.

Learn more about the friction here:

https://brainly.com/question/18754956