Chapter 1, Problem 1.22P

Interpretation Introduction

Interpretation:

Maximum height the car with mass of $\text{1200}\text{kg}$ running with speed of $89\text{km}\cdot {\text{h}}^{-1}$ can get over has to be calculated.

Concept Introduction:

Energy that is associated with a moving object is known as kinetic energy.  The energy that is possessed by an object due to its location that is related to a specific reference is known as potential energy.  Kinetic energy of any object is expressed by the equation shown below.

    $E_k=\frac{1}{2}m{v}^2$

Where $E_k$ is kinetic energy, $m$ is the mass of object, and $v$ is the velocity of object.

Potential energy for an object that is located above the ground level can be expressed by an equation given below.

    $E_p=mgh$

Where $E_p$ is the potential energy, $g$ is the proportionality constant that is equal to $9.81\text{m}\cdot {\text{s}}^{-2}$, $m$ is mass of object and $h$ is the height from ground level.

Explanation of Solution

The car has a mass of $\text{1200}\text{kg}$.  Velocity at which the car runs is $89\text{km}\cdot {\text{h}}^{-1}$.  Velocity can be converted into $\text{m}\cdot {\text{s}}^{-1}$ as shown below using conversion factors.

    $\begin{array}{c}v=89\text{km}\cdot {\text{h}}^{-1}\\ =89\text{km}\cdot {\text{h}}^{-1}\times \frac{1\text{h}}{60\min }\times \frac{\text{1}\text{min}}{\text{60}\text{s}}\times \frac{\text{1000}\text{m}}{\text{1}\text{km}}\\ =\frac{89000}{3600}\text{m}\cdot {\text{s}}^{-1}\\ =24.72\text{m}\cdot {\text{s}}^{-1}\\ =25\text{m}\cdot {\text{s}}^{-1}\end{array}$

Kinetic energy of the car is calculated as shown below.

    $\begin{array}{c}{E}_k=\frac{1}{2}m{v}^2\\ =\frac{1}{2}(1200\text{kg}){(25\text{m}\cdot {\text{s}}^{-1})}^2\\ =\text{375000}\text{kg}\cdot {\text{m}}^{\text{2}}\cdot {\text{s}}^{-2}\\ =\text{375000}\text{J}(\because 1\text{kg}\cdot {\text{m}}^{\text{2}}\cdot {\text{s}}^{-2}=1\text{J)}\\ \text{=375}\times {\text{10}}^{\text{3}}\text{J}\\ \text{=375}\text{kJ}\end{array}$

Total energy at initial level is the sum of kinetic energy and potential energy.  This can be calculated as shown.

    $\begin{array}{c}{E}_{total\text{(initial)}}=E_k+E_p\\ =375\text{kJ}+0\\ =375\text{kJ}\end{array}$

Total energy at apex is the sum of kinetic energy and potential energy.  This can be calculated as shown.

    $\begin{array}{c}{E}_{total\text{(apex)}}=E_k+E_p\\ =0+375\text{kJ}\\ =375\text{kJ}\end{array}$

Maximum height the car can reach is calculated as shown below.

    $\begin{array}{c}{E}_p=mgh\\ 375\text{kJ}=1200\text{kg}\times \text{9}\text{.8m}\cdot {\text{s}}^{-2}\times h\\ h=\frac{375\text{kJ}}{1200\text{kg}\times \text{9}\text{.8m}\cdot {\text{s}}^{-2}}\\ =\frac{375\times {10}^3\text{ kg}\cdot {\text{m}}^{\text{2}}\cdot {\text{s}}^{-2}}{11760\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}}\\ =\text{0}\text{.0318}\times {10}^3\text{m}\\ =31.8\text{m}\\ \text{=32}\text{m}\end{array}$

Therefore, the height the car can reach is calculated as $\text{32}\text{m}$.