Question

A 700 kg car going 12 m/s speeds up to 30.6 m/s in 7 seconds. How much work was done to increase its speed?

Answer 1

Answer:

Approximately $2.8 \times 10^{5} \; \rm J$ (assuming that the car is moving on level ground, and that there is no friction to hinder the motion of the car.)

Explanation:

Formula for the kinetic energy, $\rm KE$, of an object of $m$ travelling at velocity $v$:

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^{2}$.

Let $v_1$ and $v_2$ denote the velocity of this car before and after the acceleration.

Kinetic energy of this car before the acceleration:

$\begin{aligned}\text{KE}_1 &= \frac{1}{2}\, m \cdot {v_1}^{2} \\ &= \frac{1}{2} \times 700\; \rm kg \times (12\; \rm m \cdot s^{-1})^{2} \\ &\approx 5.0 \times 10^{4}\; \rm J\end{aligned}$.

Kinetic energy of this car after the acceleration:

$\begin{aligned}\text{KE}_2 &= \frac{1}{2}\, m \cdot {v_2}^{2} \\ &= \frac{1}{2} \times 700\; \rm kg \times (30.6\; \rm m \cdot s^{-1})^{2} \\ &\approx 3.3\times 10^{6}\; \rm J\end{aligned}$.

Assume that this car was travelling on level ground. Also assume that there is no friction to hinder the motion of this car. The work of the engine of this car  would be equal to the amount of $\rm KE$ that the car has gained:

$\rm KE_2 - KE_1 \approx 2.8\times 10^{5}\; \rm J$.