An object weighing 4.00 N falls from rest subject to a frictional drag force given by ,where v is the speed of the object and What terminal speed will this object approach?
A) 1.78 m/s
B) 3.42 m/s
C) 1.15 m/s
D) 2.25 m/s
E) 0.75 m/s

Question

An object weighing 4.00 N falls from rest subject to a frictional drag force given by  ,where v is the speed of the object and  What terminal speed will this object approach?
A) 1.78 m/s 
B) 3.42 m/s 
C) 1.15 m/s 
D) 2.25 m/s 
E) 0.75 m/s

Quizplus · Accepted Answer

The drag force is given by F_d = bv^2 where b = 11ea83ea_698b_8ac1_aeff_a1a4c9aaa927_TB6394_11. The net force acting on the object is F_net = ma = mg - bv^2, where m is the mass of the object and g is the acceleration due to gravity. When the object reaches terminal velocity, its speed is constant, so the net force acting on it is zero. Therefore, we have:

mg - bv_t^2 = 0

Solving for v_t, we get:

v_t = sqrt(g/b) = sqrt(4.00 N / (11ea83ea_698b_8ac1_aeff_a1a4c9aaa927_TB6394_11)) = 1.15 m/s

Therefore, the answer is choice C.

An Object Weighing 4