A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block
1 answer:
Answer:
v₁ = 0.375 m / s , x = 0.335 m
Explanation:
Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.
We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.
initial moment. Right at the start of the movement
p₀ = m v₀ + 0
final moment. Just when it comes to equilibrium
= (m + M) v₁
how the forces are internal
p₀ =p_{f}
m v₀ = (m + M) v₁
v₁ = m /m+M v₀
let's calculate
v₁ = 0.4 /(0.4 + 2.8) 3
v₁ = 0.375 m / s
Let's apply Newton's second law to the Block, to find the friction force
Y axis
N - W = 0
N = W
N = m g
where m is the mass of the block
the friction force has the formula
fr = μ N
fr = μ m g
We apply Newton's second law to slab
X axis
fr = M a
where M is the mass of the slab
μ m g = M a
a = μ g m / M
let's calculate
a = 0.15 9.8 0.4 / 2.8
a = 0.21 m / s²
With kinematics we can find the position
v²= v₀²+2 a x
as the slab is initially at rest, its initial velocity is zero
v² = 2 a x
x = v2 / 2a
let's calculate
x = 0.375²/2 0.21
x = 0.335 m
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Answer:
the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
Explanation:
You can write the KVL equations:
Top left loop:
I2(4) +(I2 +I3)(2) +I1(1) = 10
Bottom left loop:
(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10
Right loop:
(I2+I3)(2) +(I2+I3-I1)(2) = 5
In matrix form, the equations are ...
These equations have the solution ...
This means the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
_____
This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.
When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.
Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.
Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.
(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)
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It helps to be familiar with the formulas for resistors in series and parallel.