Ask your teacher a flight attendant pulls her 69 n flight bag a distance of 268 m along a level airport floor at a constant velocity. the force she exerts is 40 n at an angle of 60â° above the horizontal. (a) find the work she does on the flight bag. j (b) find the work done by the force of friction on the flight bag. j (c) find the coefficient of kinetic friction between the flight bag and the floor.

a)  W = 5360 J  b)  μ = 0.29

Explanation:

a) The work is, bold indicate vectors

         W = F. d = F d cos T

         W = 40 268 cos 60

         W = 5360 J

b) The force of friction and opposes the movement of the body and if the speed of the body is constant implies that the external force in the direction of the movement is equal to the force of friction, or which work is the same

          = -5360J

The negative sign is because the force of friction is contrary to the direction of movement

c) Let's write the work of the friction force

            = fr d cos 180

           fr = μ N

Since the body is on a horizontal surface, from Newton's second law on the Y axis

          N-W = 0

          N = W

          fr = μ W

          = - μ W d

         μ = - / W d

         μ = - (-5360) / 69 268

         μ = 0.29

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong