Chapter 5, Problem 5.1VP

Interpretation Introduction

Interpretation: The bricks kinetic energy when it is $35\text{ft}$ above the street level and its kinetic energy the instant before it hits the street surface is to be calculated.

Concept introduction: Thermodynamics involves the study of energy and its transformation from one to another form. The energy stored in the object is potential energy and kinetic energy of an object is associated with the mass and speed of an object.

To determine: The bricks kinetic energy when it is $35\text{ft}$ above the street level and the instant before it hits the street surface.

Answer to Problem 5.1VP

Solution

The bricks kinetic energy when it is $35\text{ft}$ above the street level is $\underset{\_}{150.0J}$ and the instant before it hits the street surface is $\underset{\_}{500J}$.

Explanation of Solution

Explanation

Given

Potential energy of brick is $500\text{J}$.

Since, $1\text{J}=1\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}$.

Therefore, potential energy of a brick is $500\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}$.

The height $\left(h_1\right)$ of roof edge above street level is $50\text{ft}$.

The conversion of ft to meter is done as,

$1\text{ft}=0.3048\text{m}$

Therefore the conversion of $50\text{ft}$ to meter is done as,

$\begin{array}{c}50\text{ft}=50\times 0.3048\text{m}\\ =15.24\text{m}\end{array}$

When the bricks begin to fall the potential energy converts into kinetic energy.

The mass of the brick is calculated by using the formula,

$\text{Potential}\text{energy}=m\times g\times h$

Where,

• $m$ is the mass of an object.
• $g$ is the force of gravity $\left(9.8{\text{m/s}}^{\text{2}}\right)$.
• $h$ is the vertical distance.

Substitute the values of potential energy, $g$ and $h$ in the above formula to calculate the mass of brick.

$\begin{array}{c}500\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}=m\times \left(9.8{\text{m/s}}^{\text{2}}\right)\times 15.24\text{m}\\ m=\frac{500\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}{9.8{\text{m/s}}^{\text{2}}\times 15.24\text{m}}\\ m=\frac{500\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}{149.390{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ m=3.346\text{kg}\end{array}$

The height $\left(h_2\right)$ is $35\text{ft}$.

The conversion of ft to meter is done as,

$1\text{ft}=0.3048\text{m}$

Therefore the conversion of $35\text{ft}$ to meter is done as,

$\begin{array}{c}35\text{ft}=35\times 0.3048\text{m}\\ =10.668\text{m}\end{array}$

The change in potential energy from $h_1$ to $h_2$ is calculated by using the formula.

$\text{Change}\text{in}\text{potential}\text{energy}=mg\left(h_1-h_2\right)$

Substitute the values of $m$, $g$, $h_1$ and $h_2$ to calculate the change in potential energy.

$\begin{array}{c}\text{Change}\text{in}\text{potential}\text{energy}=3.346\text{kg}\times \text{9}\text{.8}{\text{m/s}}^{\text{2}}\times \left(15.24-10.668\right)\text{m}\\ =32.7908\text{kg}{\text{m/s}}^{\text{2}}\times 4.572\text{m}\\ =149.99\text{kg}{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \approx \underset{\_}{150.0J}\end{array}$

According to the conservation of energy, the change in potential energy is equal to the change in kinetic energy.

Therefore, the change in kinetic energy is $\Delta K=\underset{\_}{150.0J}$.

When the brick is about to get hit on the surface, potential energy is zero joule. Therefore, at surface potential energy is zero that is all the potential energy gets converted to kinetic energy.

Hence, kinetic energy just before touching the surface is $\underset{\_}{500J}$.

Conclusion

The bricks kinetic energy when it is $35\text{ft}$ above the street level is $\underset{\_}{150.0J}$ and the instant before it hits the street surface is $\underset{\_}{500J}$