Question

Two-thirds of the weight of a $1500kg$ car rests on the drive wheels. What is the maximum acceleration of this car on a concrete surface?

Short answer

The maximum acceleration of this car on a concrete surface is$6.53m/s^2.$

Step-by-step solution

Given information

Two-thirds of the weight of a $1500kg$ car rests on the drive wheels.

Calculation

Mass of the car $=m_1=1500\mathrm{kg}$

Mass of the part of car on drive wheels $=m=\frac{2\times 1500}{3}=1000\mathrm{kg}$

Frictional force F will provide force for the movement of car .

Formulae used :

$$\begin{gathered} F={\mu }_k{m}_{1}g \\ \Rightarrow F=ma={\mu }_k{m}_{1}g \\ \Rightarrow a=\frac{{\mu }_k{m}_{1}g}{m}\dots \text{(1)} \end{gathered}$$

For concrete ground, ${\mu }_k=1\dots \left(2\right)$

Substituting data in equation (2) and equation (1) we get,

localid="1649266075962" $$\begin{gathered} \Rightarrow a=\frac{1\left(1000kg\right)(9.8m/s^2)}{\left(1500kg\right)} \\ \Rightarrow a=6.53\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Final answer

The maximum acceleration of this car on a concrete surface is$6.53m/s^2.$