A 3.1 kg object is initially at rest and is experiencing a net force of magnitude 4.7e^-t/2 N. As time approaches infinity, notice that the net force decays toward 0 N, which tells us that the acceleration is also dropping to 0 m/s^2 (and so the velocity is leveling off to a terminal speed). What is the terminal speed? (Ans.= 3.0 m/s)

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A 3.1 kg object is initially at rest and is experiencing a net force of magnitude 4.7e^-t/2 N. As time approaches infinity, notice that the net force decays toward 0 N, which tells us that the acceleration is also dropping to 0 m/s^2 (and so the velocity is leveling off to a terminal speed). What is the terminal speed? (Ans.= 3.0 m/s)
ent_id=_1183944 1&course_id%3D_18813 1
Problem: A 3.1 kg object is initially at rest and is experiencing a net force of magnitude
4.7e N. As time approaches infinity, notice that the net force decays toward 0 N, which tells
us that the acceleration is also dropping to 0 m/s² (and so the velocity is leveling off to a
terminal speed). What is this terminal speed? (ANS. = 3.0 m/s)
(A) Pictorial
Representation
(B) Physical
Representation
(C) Math
Representation
and Solution
2:
Transcribed Image Text:ent_id=_1183944 1&course_id%3D_18813 1 Problem: A 3.1 kg object is initially at rest and is experiencing a net force of magnitude 4.7e N. As time approaches infinity, notice that the net force decays toward 0 N, which tells us that the acceleration is also dropping to 0 m/s² (and so the velocity is leveling off to a terminal speed). What is this terminal speed? (ANS. = 3.0 m/s) (A) Pictorial Representation (B) Physical Representation (C) Math Representation and Solution 2:
Expert Solution
Step 1

Solution:

Let net force acting on the body be dF

Since it is given that,

dF=4.7e-t/2N

Now, 

since, 

dF=ma=mdvdt

Then

mdvdt=4.7×e-t2mdv=4.7e-t2

As velocity v varies from 0 to terminal velocity (v), then time varies from 0 to , therefore,

m0vdv=4.70e-t2dtm[v-0]=4.7×e-t/20-12mv=4.7×-2 [0-1]3.1×v=4.7×2v=3.03m/s 3m/s

Terminal velocity on the mass=3m/s

 

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