Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as

By putting the values

ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation


150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
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This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
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