# Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force \vec{F} The magnitude

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force $\stackrel{\to }{F}$ The magnitude of the tension in the string between blocks B and C is T=3.00N. Assume that each block has mass m=0.400kg.
What is the magnitude F of the force?
What is the tension in the string between block A and block B?
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Ezra Herbert
The magnitude of F of the force
T=ma
$3.00N=0.400kg\cdot a$
thus

But this is the total acceleration, now for each individual block the

now one can find F by using
$a=\frac{F}{3m}$ and solving for F
$F=a\cdot 3m$
$F=4.5N$
For the Tension between block A and B, all we do is
T=ma, knowing that for each block.
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Jeffrey Jordon

${T}_{2}=3N$

M=0.4 kg

${T}_{1}=0.4a$  -(1)

${T}_{2}-{T}_{1}=0.4a$ -(2)

$F-{T}_{2}=0.4a$ -(3)

$a=3.75m/{s}^{2},{T}_{AB}=1.5N$