The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.
Answer:
The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v
Explanation:
Solution
Given that:
V+ = 20v
Re = 2kΩ
Rc = 1kΩ
Now we will amke use of the method KVL in the loop.
= - Ve + IE . Re + VEB + VB = 0
Thus
IE = V+ -VEB -VB/Re
Which gives us the following:
IE = 20-0.7 - 10/2k
= 9.3/2k
so, IE = 4.65 mA
IB = IE/β +1 = 4.65 m /101
Thus,
IB = 0.046039 mA
IB = 46.039μA
IC =βIB
Now,
IC = 100 * 0.046039
IC is 4.6039 mA
Now,
VB = 10v
VE = VB + VEB
= 10 +0.7 = 10.7 v
So,
Vc =Ic . Rc = 4.6039 * 1k
=4.6039 v
Finally, this is the table summary from calculations carried out.
Summary Table
Parameters IE IC IB VE VB Vc
Unit mA mA μA V V V
Value 4.65 4.6039 46.039 10.7 10 4.6039
Answer:
The speed at point B is 5.33 m/s
The normal force at point B is 694 N
Explanation:
The length of the spring when the collar is in point A is equal to:

The length in point B is:
lB=0.2+0.2=0.4 m
The equation of conservation of energy is:
(eq. 1)
Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2
in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2
Replacing in eq. 1:

Replacing values and clearing vB:
vB = 5.33 m/s
The balance forces acting in point B is:
Fc-NB-Fs=0

Replacing values and clearing NB:
NB = 694 N