**Calculate the work done by the customer on the cart as she drives down a****50m****long slide.****What is the net work done on the cart? Explain.****The customer goes down the next slide, moving horizontally and keeping the same speed as earlier. If the friction force doesn’t alter, would the customer’s devoted force be more, less, or unchanged?****What do you say regarding the work done on the cart by the customer?**

This problem aims to find the **work done** by the **customer** on the **cart** as she slides down the **hall.** 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.

\[ F_s = F_x = F\cos \theta \space s \]

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

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

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

## Expert Answer

**Part a:**

We are given the following **information:**

**Magnitude** of **force** $F = 35 N$,

The **angle** at which the **force** occurs $\theta = 25 $ and,

The **displacement** $\bigtriangleup s = 50 m$.

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

\[ W_{customer} = F \times s = (F\cos \theta) \times \bigtriangleup s\]

\[ W = (35.0 N)(50.0 m)\cos 25\]

\[W=1.59\times 10^3\space J\]

**Part b:**

Since the **cart** moves at a **constant speed,**

\[ F_x – f=0 \implies f=+F\cos25 \]

Where $f$ is the **work done** by **friction.**

\[ W_f=fx\cos 180^{\circ}\]

\[=-fx\]

\[=-F\cos 35\times x\]

\[=-1586J\]

Since $W_{net}=W_s+W_f $

So $W_{net}=0$, as the **speed** doesn’t **change.**

**Part c:**

Since the cart stays at a **constant velocity,** the **force** exerted on the cart will be equal to the **friction force** as it is now completely **horizontal** to the surface. Thus the net **work** **done** on the cart will be equal to the change in **kinetic energy** being generated due to the **change** in position.

\[W_{net}=\bigtriangleup K.E.\]

Since the **speed** doesn’t change,

\[W_{net}=0\]

We know that the net **work done** $W_{net}$ is the sum of **frictionless** work $W_s$ and work under the **force** of **friction** $W_f$, so:

\[W_{net}=W_s+W_f \]

\[W_s=-W_f \]

Also, $F_{net}=-f$, which says that the **friction** is smaller when the customer pushes the cart **horizontally.**

## Numerical Result

**Part a:** $W=1.59\times 10^3\space J$

**Part b:** $W_{net}=0$

**Part c:** $W_s=-W_f$

## 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** $S=50m$,

\[ W=F\times S\]

\[=250\times50\]

\[=1250\space J\]

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