**The particle moves from the origin O to a final position with the coordinates as x=4.65m and y=4.65m, which is also represented in the following figure.**

**Figure 1**

**Find work done by F along OAC****Find work done by F along OBC****Find work done by F along OC****Is F conservative or non-conservative?**

This problem aims to find the **work done** by the **particle** moving in the **xy**Â plane as it moves to the new position with the given coordinates**.** The concepts required for this problem are related to **basic physics,** which includes** work done on a body** and** friction force.**

The concept of **work done** comes as the **dot product** of the **horizontal** component of the **force** with the **direction** of the **displacement** along with the **value of the displacement**.

\[ F_s = F_x = Fcos \theta \space s \]

The **component** that is responsible for the **movement** of the object is $Fcos\theta$, where $\theta$ is the **angle** between the **force** $F$ and the **displacement** **vector** $s$.

Mathematically,Â **Work done** is a **scalar** quantity and is **expressed** as:

\[ W = F \times s = (Fcos \theta) \times s \]

Where $W=$ **work,** $F=$ **force** exerted.

## Expert Answer

**Part A: **

**Work done by $F$ along $OAC$**

We are given the following **information:**

**Force** $F = (2y i + x^2 j) N$,

The **displacement** in the direction of $x = 4.65 m$ and

The **displacement** in the direction of $y = 4.65 m$.

To calculate the **work done, according to the given figure** we are going to use the **formula:**

\[W=\dfrac {1}{2} \times\ x \times y\]

\[W=\dfrac {1}{2} \times\ 4.65 \times 4.65\]

\[W=\dfrac {1}{2} \times\ 21.6225\]

\[W= 10.811 \space J\]

**Part B:**

**Work done by $F$ along $OBC$**

**Force** $F = (2y i + x^2 j) N$,

The **displacement** in the direction of $x = 4.65 m$ and

The **displacement** in the direction of $y = 4.65 m$.

\[W=\dfrac{1}{2} \times\ x \times y\]

\[W=\dfrac{1}{2} \times\ 4.65 \times 4.65 \]

\[W=\dfrac{1}{2} \times\ 21.6225 \]

\[W=10.811 \space J\]

**Part C:**

**Work done by $F$ along $OC$**

We are given the following **information:**

**Force** $F = (2y i + x^2 j) N$,

The **displacement** in the direction of $x = 4.65 m$ and

The **displacement** in the direction of $y = 4.65 m$.

The **position of particle** at the **point** $C = (4.65 i+4.65 j)$

To calculate the **work done** we are going to use the **formula:**

\[W_{particle}=F \times s = (2y i + x^2 j)(4.65 i+4.65 j)\]

\[W_{particle}=(2(4.65) i + (4.65)^2 j) (4.65 i+4.65 j)\]

\[W_{particle}=143.78\space J\]

**Part D: **

**Non-conservative Force**

## Numerical Result

**Part A:** $10.811\space J$

**Part B:** $10.811\space J$

**Part C:** $143.78\space J$

**Part D: Non-conservative Force**

## Example

Find the **work done** in driving a cart via a **distance** of $50 m$ **against** the **force of friction** of $250N$. Also, comment on the kind of **work done.**

We are **given:**

The **Force** exerted $F=250N$

**Displacement** $S=50m$

\[ W=F\times S\]

\[W=250\times50\]

\[W=1250\space J\]

Note that the **work**Â **done** here is **negative**.

*Image/Mathematical drawings are created in Geogebra.*