3. In this course, we have mostly avoided dealing with air drag forces (which va

Question

3. In this course, we have mostly avoided dealing with air drag forces (which vary with velocity and lead to differential equations). But we can get some sense via an average drag force, FDayg in some cases. Heres a situation where we do that-and to illustrate the flexibility of the physics "toolkit" you now have, for each item, you're asked to solve it two different ways-once using mechanical energy analysis, again via Newton's Laws and kinematics. In two separate trials (A and B), the same ball m 0.687 kg) is dropped from rest from a high bridge-from a height of exactly 100 m above the level ground below. Motion sensors placed on the ground reveal the impact speed. Trial A-calm air (no wind): The ball's impact speed is 43.0 m/s. Its landing point is directly below the drop point. Trial B-steady wind, blowing west to east: The ball's impact speed is 43.5 m/s. Its landing point is 12.9 m directly east of the landing point of Trial A. And cameras reveal a linear (not parabolic) path from release to impact In neither of the above trials is the ball a true projectile. So, assuming that the freefall acceleration of a true projectile (i.e. without any air drag) sg 9.80 m/s2, do each of these calculations two different ways (energy and kinematics). a. For trial A, find the magnitude of the average vertical drag force, FDyang, exerted by the air on the ball b. For trial B-assuming the same average vertical drag force as in part a-find the magnitude of the average horizontal force, FDaexerted by the air on the ball. xavg

Explanation / Answer

a. using energy method,

net work done = change in KE

Work done by gravity + work done by drag force= change iN KE

(0.687 x 9.8 x 100 ) + ( - 100 F) = (0.687)(43^2 - 0^2) / 2

673.26 - 100 F = 635.13

F = 0.38 N .........Ans

Kinetimatics:

vf^2 - vi^2 = 2 a d

43^2 - 0 = 2 (a) (100)

a = 9.245 m/s^2

and Fnet = m g - F = m a

0.687 x 9.8 - F = 0.687 x 9.245

F = 0.38 N

(B) Energy Method:

Work done by horizontal drag force = difference in KE

12.9 Fx = 0.687(43.5^2 - 43^2) /2

Fx = 1.15 N  

Kinetmetic method:

in hoizontal,

v0 = 0

vf = sqrt(43.5^2 - 43^2) = 6.576 m/s

vf^2 - vi^2 = 2 a d

6.576^2 - 0^2 = 2(a)(12.9)

a =1.676 m/s^2

in horizontal, F = m a

Fx = 0.687 x 1.676 = 1.15 N

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