**If the coefficient of kinetic friction of their bodies with the floor is 0.250, how far do they slide?**

The question aims to understand **newton’s law** of motion, the **law** of **conservation,** and the **equations** of **kinematics.**

**Newton’s** law of motion states that the **acceleration** of any object relies on **two variables,** the **mass** of the object and the **net force** acting on the object. The **acceleration** of any object is **directly** proportional to the **force acting** on it and is **inversely** proportional to the **mass** of the object.

**principle**that does not

**change**and states a certain

**property**in the course of

**time**within an isolated

**physical**system is called

**conservation law.**Its equation is given as:

**potential**energy and K is the

**kinetic**energy.

**motion**of objects using

**diagrams, words, graphs, numbers**and

**equations**is described as

**Kinematics.**The aim of

**studying**the kinematics is to design

**sophisticated**mental models that help in

**describing**the motions of

**physical**objects.

## Expert Answer

In the **question,** it is given that:

Stuntman has a mass of $(m_s) \space= \space 80.0kg$.

Movie’s villain has a mass of $(m_v)= \space 80.0kg$.

The **distance** between the floor and window is $h= \space 5.0m$.

**Part a**

Before the **collision** of the stunt man, the initial **velocity** and the final **height** is $0$, therefore the $K.E = P.E$.

\[ \dfrac{1}{2}m_sv_2^2 = m_sgh\]

\[v_2 = \sqrt{2gh}\]

Therefore the **speed** $(v_2)$ becomes $\sqrt{2gh}$.

Using the **law** of conservation, the **speed** after the collision can be calculated as:

\[v_sv_2= (m_s+ m_v) .v_3\]

Making $v_3$ the subject:

\[v_3 = \dfrac{m_s}{m_s+ m_v} v_2\]

Plugging $v_2$ back in:

\[v_3= \dfrac{m_s}{m_s+ m_v} \sqrt{2gh}\]

Plugging the values and **solving** for $v_3$:

\[ v_3 = \dfrac{80}{80+ 70} \sqrt{2(9.8)(5.0)} \]

\[ v_3 = \dfrac{80}{150}. 9.89 \]

\[v_3 = 5.28 m/s\]

**Part b**

The **coefficient** of **kinetic** friction of their bodies with the floor is $(\mu_k) = 0.250$

Using **Newton’s** 2nd law:

\[ (m_s + m_v)a = – \mu_k (m_s + m_v)g \]

**Acceleration** comes out to be:

\[ a = – \mu_kg \]

Using the **Kinematics** formula:

\[ v_4^2 – v_3^2 = 2a \Delta x \]

\[ \Delta x = \dfrac{v_4^2 – v_3^2}{2a} \]

Inserting the **acceleration** $a$ and putting **final velocity** $v_4$ equals $0$:

\[ = \dfrac{0 – (v_3)^2}{ -2 \mu_kg} \]

\[ = \dfrac{(v_3)^2}{2 \mu_kg} \]

\[ = \dfrac{(5.28)^2}{2(0.250)(9.8)} \]

\[\Delta x = 5.49 m\]

## Numerical Answer

**Part a: **Entwined foes start to **slide** across the floor with the **speed** of $5.28 m/s$

**Part b: **With **kinetic** friction of 0.250 of their **bodies** with the **floor,** the sliding **distance** is $5.49m$

## Example:

On the runway, an airplane **accelerates** at $3.20 m/s^2$ for $32.8s$ until it **finally** lifts off the ground. Find the distance **covered** before takeoff.

Given that **acceleration** $a=3.2 m/s^2$

**Time** $t=32.8s$

Initial **velocity** $v_i= 0 m/s$

**Distance** $d$ can be found as:

\[ d = vi*t + 0.5*a*t^2 \]

\[ d = (0)*(32.8) + 0.5*(3.2)*(32.8)^2 \]

\[d = 1720m\]