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Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force 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 tab in the string between block A and block B?

This article aims to find the tension in the string between two blocks $ A $ and $ B $. This article uses the concept of how to find the tension in the string. Tension in physics is the force developed in a rope, string, or cable when an applied force stretches it. Tension acts along the length of the rope in the opposite direction to the force acting on it. Tension can sometimes be referred to as stress, strain, or tension.

The formula for the tension in a string is given as:

\[ T = ma \]

Expert Answer

Given data

\[T = 3.00\: N \]

\[m = 0.400 \: kg \]

The magnitude $ F $ of the force is given by:

\[ T = m a \]

\[ 3.00 = ( 0.400 ) a \]

\[ a = \dfrac { 3 }{ 0.400 } \]

\[a = 7.5 \dfrac {m}{s^{2}} \]

This is the total acceleration; acceleration for the individual block is:

\[ a = \dfrac {7.5}{2} = 3.75 \dfrac {m}{s^{2}} \]

Force $F $ can be found by using:

\[ a = \dfrac {F}{3m} \]

\[F = 3am \]

\[F = 3 (3.75)(0.400 ) \]

\[ F = 4.5\:N \]

For the tension between block $ A $ and $ B $:

\[ T = ma \]

\[T = (0.400\:kg) (3.75 \dfrac {m}{s^{2}}) \]

\[T = 1.5 \: N \]

The tension for each block is $1.5 \: N $.

Numerical Result

The tension for each block is $1.5 \: N $.

Example

Three identical blocks connected by ideal strings are pulled along a frictionless horizontal surface by a horizontal force $ F $.

Magnitude of the tension in string between blocks $ B $ and $ C $ is $ T=5.00\:N $. Assume that each block has a mass of $ m=0.500 \:kg$.

-What is the magnitude of the $ F $ force?

-What is the tension in the string between block $ A $ and block $ B $?

Solution

Given data

\[T = 5.00\: N \]

\[m = 0.500 \: kg \]

The magnitude $ F $ of the force is given by:

\[ T = m a \]

\[ 5.00 = ( 0.500 ) a \]

\[ a = \dfrac { 5 }{ 0.500 } \]

\[a = 10 \dfrac { m }{s ^ { 2 }} \]

This is the total acceleration; acceleration for the individual block is:

\[ a = \dfrac { 10 }{ 5 } = 2 \dfrac { m }{ s ^ { 2 }} \]

Force $F $ can be found by using:

\[ a = \dfrac { F }{ 3 m } \]

\[F = 3 a m \]

\[F = 3 ( 2 )( 0.500 ) \]

\[ F = 3 \:N \]

For the tension between block $ A $ and $ B $:

\[ T = ma \]

\[T = ( 0.500\:kg ) ( 2 \dfrac {m}{s ^ { 2 }} ) \]

\[T = 1.0 \: N \]

The tension for each block is $ 1.0 \:N $.

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