The** aim of this question** is to develop an understanding of the basic concepts of **work done** and **resultant force.**

The **work done** is a** scalar quantity** defined as the** amount of energy** dispensed whenever a **forcing agent** moves a body along **some distance** in the direction of the force. Mathematically, it’s defined as the **dot product of force and displacement**.

\[ W \ = \ \vec{ F } . \ \vec{ d } \]

Where W is the **work done**, F is the **average force** and d is the **displacement**. If force and displacement are **colinear,** then the above equation reduces to:

\[ W \ = \ | \vec{ F } | \times | \vec{ d } | \]

Where $ | \vec{ F } | $ and $ | \vec{ d } | $ are the **magnitudes** of force and displacement.

Whenever **two or more forces** act on a body, the **body moves** in the direction of the net force or **resultant force.** Net force or resultant force is the **vector sum of all the forces** acting on the said body. The net force can be c**alculated using** vector addition methods such as a **head-to-tail rule** or **polar coordinate** addition or **complex addition** etc.

## Expert Answer

**Given that:**

\[ \text{ Work Done } = \ W \ = \ 80 \ J \]

\[ \text{ Distance Covered } = \ d \ = \ 0.2 \ m \]

From the definition of **work done**, we can find the **average force** on one spring during this movement by utilizing the following formula:

\[ \text{ Work Done } = \text{ Average Force } \times \text{ Distance Covered } \]

\[ W \ = \ F \times \ d \]

\[ \Rightarrow F \ = \ \dfrac{ W }{ d } \ … \ …\ … \ ( 1 ) \]

**Substituting given values:**

\[ F \ = \ \dfrac{ 80 \ J }{ 0.2 \ m } \]

\[ \Rightarrow F \ = \ 400 \ N \]

Since there are** two springs**, so the **net force needed** to press both the springs by a distance of 0.2 m **will be twice**:

\[ F_{ net } \ = \ 2 \times 400 \ N \]

\[ \Rightarrow F_{ net } \ = \ 800 \ N \]

## Numerical Result

\[ F_{ net } \ = \ 800 \ N \]

## Example

Given the **same platform**, how much **force** will be needed to push the platform **by a distance of 0.400 m** from the uncompressed position?

**Recall equation (1):**

\[ \Rightarrow F \ = \ \dfrac{ W }{ d } \]

**Substituting given values:**

\[ F \ = \ \dfrac{ 80 \ J }{ 0.4 \ m } \]

\[ \Rightarrow F \ = \ 200 \ N \]

Since **there are two springs**, so the **net force needed** to press both the springs by a distance of 0.4 m **will be twice**:

\[ F_{ net } \ = \ 2 \times 200 \ N \]

\[ \Rightarrow F_{ net } \ = \ 400 \ N \]