This **article aims**Â to find the tension in the **string between two blocks**Â $ A $ and $ B $. This article uses the**Â concept**Â of how to find the **tension in the string.**Â **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**Â can sometimes be referred to as **stress, strain, or tension**.

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

\[ T = ma \]

**Expert Answer**

**Given data **

\[T = 3.00\: N \]

\[m = 0.400 \: kg \]

The **magnitude $ F $ of the force**Â 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**Â individual block**Â 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**Â $ A $ and $ B $:

\[ T = ma \]

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

\[T = 1.5 \: N \]

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

**Numerical Result**

The **tension for each block**Â 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**Â 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**Â individual block**Â 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**Â $ A $ and $ B $:

\[ T = ma \]

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

\[T = 1.0 \: N \]

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