**Find the total mass occupied by the block of ice.****If the worker quits moving at the end of**Â**5s****, how long does the block move in the next****5s?**

This problem aims to familiarize us with the **applied force** and the **acceleration** of a moving **body.** The concepts required to solve this problem are from **basic applied physics** which include the **sum** of **applied force, instantaneous speed,** and **newtons law** of **motion.**

Let’s first look at **instantaneous velocity,** which notifies us how fast an object is **moving** at a particular **instance** of **time,** simply named **velocity.** It is basically the average velocity **between** two points. The only **difference** lies in the limit that the time between the **two circumstances** closes to **zero.**

\[ \vec{v} = \dfrac{x(t_2) – x(t_1)}{t_2 – t_1} \]

## Expert Answer

We are given the following **information:**

A **horizontal force** $F_x = 80.0 \space N$,

The **distance** the car travels from **rest** $s = x – x_0 = 11.0 \space m$,

**Part a:**

First, we are going to find the **acceleration** using the **newton equation** of **motion:**

\[ sÂ = v_it + \dfrac{a_x t^2}{2} \]

Since the car **starts** from **rest,** so $v_i = 0$:

\[ 11Â = 0 + \dfrac{a_x \times 25}{2} \]

\[ 22Â = a_x\times 25 \]

\[ a_x = \dfrac{22}{25} \]

\[ a_x = 0.88 m/s^2 \]

Using the **first equation** of **motion,** we can find the **mass** of the object moving with an **acceleration** of $a = 0.88 m/s^2$:

\[ F_x = ma_x \]

\[ m = \dfrac{F_x}{a_x} \]

\[ m = \dfrac{80.0 N}{0.880 m/s^2} \]

\[ m = 90.9 \space kg \]

**Part b:**

At the end of $5.00 s$, the **worker** stops **pushing** the **block** of ice, which means its **velocity** remains **constant** as the **force** becomes **zero.** We can find that **velocity** using:

\[ v_x = a_x \times t \]

\[ v_x = (0.88 m/s^2)(5.00 s) \]

\[ v_x=4.4 m/s\]

So, after $5.00 s$, the **block** of **ice** moves with a constant **velocity** of $v_x = 4.4 m/s$.

Now to find the **distance** the block **covers,** we can use the **distance formula:**

\[ s=v_x\times t\]

\[ s=(4.4 m/s)(5.00 s)\]

\[s=22\space m\]

## Numerical Result

The **mass** of the **block** of ice is: $m = 90.9\space kg$.

The **distance** the **block** covers is $s = 22\space m$.

## Example

A **worker drives** a box with $12.3 kg$ on a **horizontal** surface of $3.10 m/s$. The coefÂficients of **kinetic** and **static friction** are $0.280$ and $0.480$, respectively. What force must the **worker** use to sustain the **motion** of the box?

Let’s set the **coordinate** so that the **motion** is in the **direction** of the $x$-axis. Thus **Newton’s second law** in **scalar** form appears like this:

\[F-f=0\]

\[N-mg=0\]

We know that **frictional force** $f=\mu k\space N$, we will get $f=\mu kmg$. Since the body is **moving,** we use the **coefficient** of **kinetic friction** $\mu k$.

Then we can **rewrite** the **equation** as:

\[F-\mu kmg=0\]

Solving for **force:**

\[F=\mu kmg\]

**Substituting** the values:

\[F=0.280\times 12.3\times 9.8\]

\[F=33.8\space N\]