Chapter 3, Problem 50P

To determine

The additional power needed to achieve the acceleration.

The additional power needed to achieve the acceleration when the total mass of the car is $700\text{kg}$.

Explanation of Solution

Given:

Mass of the car $\left(m_{\text{car}}\right)$ is $1400\text{kg}$.

Initial speed of the car $\left(V_1\right)$ is $70\text{km}/\text{h}$.

Final speed of the car $\left(V_2\right)$ is $100\text{km}/\text{h}$.

Change in time interval of the car $\left(\Delta t\right)$ is $5\text{s}$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Calculate the additional power needed to achieve the acceleration $\left({\dot{W}}_{\text{in}}\right)$ using the energy balance equation.

  $\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}-0=d{E}_{\text{system}}/dt\end{array}$

  $\begin{array}{c}{\dot{E}}_{\text{in}}=d{E}_{\text{system}}/dt\\ =\frac{\Delta E_{\text{sys}}}{\Delta t}\end{array}$

  ${\dot{W}}_{\text{in}}=\frac{\Delta KE}{\Delta t}$

  $\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{m\left(V_2^2-V_1^2\right)}{2\Delta t}\\ =\frac{\left(1400\text{kg}\right)\left[{\left(110\text{km}/\text{h}\right)}^2-{\left(70\text{km}/\text{h}\right)}^2\right]}{2\left(5\text{s}\right)}\\ =\frac{\left(1400\text{kg}\right)\left[{\left(110\times \frac{5}{18}\text{m}/\text{s}\right)}^2-{\left(70\times \frac{5}{18}\text{m}/\text{s}\right)}^2\right]}{2\left(5\text{s}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =77.8\text{kJ}/\text{s}\left(\frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =\text{77}\text{.8}\text{kW}\end{array}$

Thus, the additional power needed to achieve the acceleration is $\underset{\_}{77.8\text{kW}}$.

Calculate the additional power needed to achieve the acceleration $\left({\dot{W}}_{\text{in}}\right)$ when the total mass of the car is $700\text{kg}$.

  ${\dot{W}}_{\text{in}}=\frac{m\left(V_2^2-V_1^2\right)}{2\Delta t}$

  $\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\left(700\text{kg}\right)\left[{\left(110\text{km}/\text{h}\right)}^2-{\left(70\text{km}/\text{h}\right)}^2\right]}{2\left(5\text{s}\right)}\\ =\frac{\left(700\text{kg}\right)\left[{\left(110\times \frac{5}{18}\text{m}/\text{s}\right)}^2-{\left(70\times \frac{5}{18}\text{m}/\text{s}\right)}^2\right]}{2\left(5\text{s}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =38.9\text{kJ}/\text{s}\left(\frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =\text{38}\text{.9}\text{kW}\end{array}$

Thus, the additional power needed to achieve the acceleration is $\underset{\_}{\text{38}\text{.9}\text{kW}}$.