Answer:
1.32 m/s² & 149.4 N
Step-by-step explanation:
To solve this problem, we'll use Newton's second law, which states that the net force acting on an object is equal to the object's mass times its acceleration (F=ma). We'll apply this law to both blocks separately.
First, let's find the force of friction acting on the blocks. We know the coefficient of friction is 0.196, and the normal force (the force perpendicular to the surface) acting on each block is equal to its weight. So the force of friction on block 1 (m₁) is:
f₁ = μ₁N₁ = 0.196(m₁g)
where g is the acceleration due to gravity (9.8 m/s²). Similarly, the force of friction on block 2 (m₂) is:
f₂ = μ₂N₂ = 0.196(m₂g)
Next, let's apply Newton's second law to each block. For block 1:
F - f₁ = m₁a
where F is the tension in the string. For block 2:
F - f₂ - 100 = m₂a
We have two equations and two unknowns (a and F), so we can solve for both. First, let's eliminate F by adding the two equations together:
(F - f₁) + (F - f₂ - 100) = (m₁ + m₂)a
2F - f₁ - f₂ - 100 = (m₁ + m₂)a
Now we can solve for a:
a = (2F - f₁ - f₂ - 100)/(m₁ + m₂)
To find F, we'll plug in the known values and solve for F:
a = (2F - f₁ - f₂ - 100)/(m₁ + m₂)
a(m₁ + m₂) = 2F - f₁ - f₂ - 100
2F = a(m₁ + m₂) + f₁ + f₂ + 100
F = (a(m₁ + m₂) + f₁ + f₂ + 100)/2
Now we just need to plug in the values and solve:
m₁ = 11.1 kg
m₂ = 25.8 kg
F = ?
f₁ = 0.196(m₁g) = 21.68 N
f₂ = 0.196(m₂g) = 50.78 N
μ₁ = μ₂ = 0.196
g = 9.8 m/s²
F_applied = 100 N
a = (2F - f₁ - f₂ - 100)/(m₁ + m₂)
We know that a = F_net/(m₁ + m₂), where F_net = F_applied - f₁ - f₂, so we can substitute:
a = (F_applied - f₁ - f₂)/(m₁ + m₂)
a = (100 N - 21.68 N - 50.78 N)/(11.1 kg + 25.8 kg) = 1.32 m/s²
Now we can solve for F:
F = (a(m₁ + m₂) + f₁ + f₂ + 100)/2
F = [(1.32 m/s²)(11.1 kg + 25.8 kg) + 21.68 N + 50.78 N + 100 N]/2 = 149.4 N
So the acceleration of the system is 1.32 m/s², and the tension in the string is 149.4 N.