(Force is assumed to be constant) 3. In this course, we have mostly avoided deal

Question

(Force is assumed to be constant)

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, Fons in some cases. Here's 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 waxs-once using mechanical energy analysis, again via Newton's Laws and kinematics. In two separate trials (A and B), the same ball (m = 0687 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 lincar (not paraholic) path from relcase to impact. eefall acceleration of a true projectile exerted by the air on the ball. In neither of the above trials is the ball a true projectile. So, assuming that the f (ie, without any air drag) is g 9.80 m/s, a. For trial A, find the magnitude of the average vertical drag forue, Fa b. For trial B-assuming the average same average vertical drag force as in part a-find the magnitude of the average horizontal force, Fg, cxerted by the air on the bal

Explanation / Answer

a)
Total initial energy = mgh
= 0.987 x 9.8 x 100 = 673.26 J
Total final energy = 1/2 mv2
= 0.5 x 0.687 x (43)2
= 635.1315 J

Difference in energy = Work done against the drag force
= 635.1315 - 673.26
= - 38.1285 J
Work done = Fdrag x distance
Fdrag = 38.1285/100 = 0.38 N

b)
Net acceleration alog the vertical direction = g - Fdrag/m
= 9.8 - 0.38/0.687
= 9.245 m/s2.

Time taken by the mass to reach a velocity 43 m/s is,
t = v/a = 43/9.245
= 4.65 s

Now consider the horizontal motion.
Initial velocity = 0
Distance traveled, s = 12.9 m
Time taken = 4.65 s
Using the formula, s = ut + 1/2 at2,
a = 2s/t2
= [2 x 12.9] / 4.652
= 1.193 m/s2.

Horizontal force = Horizontal acceleration x mass
= 1.193 x 0.687
= 0.82 N

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