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

Question

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force

\vec{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?

Answer 1

The magnitude of F of the force
T=ma

3.00 N = 0.400 k g \cdot a

thus

a = 7.5 \frac{m}{s^{2}}

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

a = \frac{7.5}{2} = 3.75 \frac{m}{s^{2}}

now one can find F by using

a = \frac{F}{3 m}

and solving for F

F = a \cdot 3 m

F = 4.5 N

For the Tension between block A and B, all we do is
T=ma, knowing that

a = 3.75 \frac{m}{s^{2}}

for each block.

T = (0.400 k g) (3.75 \frac{m}{s^{2}}) = 1.5 N

Answer 2

$T_{2} = 3 N$

M=0.4 kg

$T_{1} = 0.4 a$ -(1)

$T_{2} - T_{1} = 0.4 a$ -(2)

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

$a = 3.75 m / s^{2}, T_{A B} = 1.5 N$

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

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