**– Calculate the velocity of the rock when the string perishes via the vertical standing.****– Find the net tension in the line when it completes an angle of**$45$**with the vertical position.****– If the string passes through a vertical position, what will be the tension in the string?**

This problem aims to find the **tension** in a **string** as it passes through **vertical** space. The concept required to solve this problem is related to **tension** and **acceleration** and the **law** of **thermodynamics** which includes **kinetic energy** and **potential energy.**

Let’s first look into the phrase **tension,** which is a **force** acting along the **size** of a **medium,** particularly a **force** held by an **elastic** substance, like a **rope** or a **string.** It is an **action-reaction** couple of **forces** functioning at each end of the elastic **medium.**

The **tension** $(T)$ on an **elastic** substance having **mass** $(m)$ is equivalent to the mass time **gravitational force** $(g)$ summed to the **mass** times **acceleration** $(a)$ expression. This whole definition can be summed up as:

\[T=mg+ma\]

The **first law** of **thermodynamics,** which is most commonly known as the **law** of **conservation** of **energy,** states that the **energy** possessed by a substance can neither be **created** nor can be **eliminated** but can **transform** from one form to another. The common **expression** for this is:

\[K.E=P.E\]

## Expert Answer

Before starting the **solution,** we will list out the given **variables,**

The **length** of **string** $l=0.8 m$,

**Mass** of the rock $m=0.12kg$,

And the **angle** with the **vertical position** $\theta=45^{\circ}$.

**Part a:**

For this part, we will have to find the **rate** at which the **rock attached** to the string passes through the **vertical position.** According to the first rule of **thermodynamics:**

\[\text{Potential energy}=\text{Kinetic energy}\]

**Potential energy** = $mgl\times (1-\cos\theta)$.

**Kinetic** **energy** = $\dfrac{1}{2}mv^2$.

Plugging them in:

\[m\times g\times l\times (1-\cos\theta)=\dfrac{1}{2}\times m\times v^2\]

**Rearranging** the above **equation** for calculating $v$:

\[v^2=2\times g\times l\times (1-\cos\theta)\]

\[=2\times 9.8\times 10.8\times (1-\cos(45))=4.597\]

We get,

\[v=\sqrt{4.597}=2.1 ms^{-1}\]

**Part b:**

Finding the **tension** in the line when it **completes** an **angle** of $45$ using the **first condition** of **equilibrium:**

\[\sum{Forces}=0\]

\[T-m\times g\times \cos\theta=0\]

\[T=m\times g\times \cos\theta\]

\[T=0.12\times 9.81\times \cos(45)\]

\[T=0.83 N\]

**Part c:**

Finally, the **tension** in the **string** as it crosses through the **vertical position** can be found by the **first condition** of **equilibrium:**

\[T – m\times g = \dfrac{(m\times v^2)}{r}\]

\[T=mg(1+2(1-\cos\theta))\]

\[T=0.981\times 0.12 (1+ 2(1-\cos(45)))\]

\[T=1.867 N\space or\space 1.9 N\]

## Numerical Result

**Part a:** $v=2.1 ms^{-1}$

**Part b:** $T=0.83 N$

**Part c:** $T=1.867 N\space or\space 1.9 N$

## Example

A $10 kg$ **rock** is hanging from a **rope.** Find the **tension** in the rope, if the rock:

- – experiences
**zero acceleration?** - – has an
**upward acceleration**of $5 m/s^2$. - – has a
**downward acceleration**of $-5 m/s^2$.

To calculate **tension:**

\[T=mg+ma\]

**Substituting** the values:

\[T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (0)\]

\[T=108 N\]

With an **upward acceleration** of $5 m/s^2$, we get:

\[T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (5 m/s^2)\]

\[T=148 N\]

With a **downward acceleration** of $-5 m/s^2$, we get:

\[T=(10 kg)\times (9.8 m/s^2)+(10 kg)\times (-5 m/s^2)\]

\[T=48N\]