Question

Calculate the velocity of a truck with kinetic energy 2.7kJ and mass 600 kg.

Answer 1

Answer:

Approximately $3.0\; {\rm m\cdot s^{-1}}$.

Explanation:

When an object of mass $m$ travels at a speed of $v$, the kinetic energy $(\text{KE})$ of the object would be:

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

In this question, it is given that $(\text{KE}) = 2.7\; {\rm kJ}$ and $m = 600\; {\rm kg}$. Rearrange the equation above to find velocity $v$:

$\begin{aligned}v &= \sqrt{\frac{2\, (\text{KE})}{m}}\end{aligned}$.

Before substituting in the values, make sure that all quantities are in standard units:

• Energy should be in joules: $(\text{KE}) = 2.7\; {\rm kJ} = 2.7 \times 10^{3}\; {\rm J}$.
• Mass should be in kilograms: $m = 600\; {\rm kg}$.

Also note that $1\; {\rm J} = 1\; {\rm N\cdot m} = 1\; {\rm kg\cdot m^{2}\cdot s^{-2}}$, so the kinetic energy in this question can be expressed as $(\text{KE}) = 2.7 \times 10^{3}\; {\rm kg \cdot m^{2}\cdot s^{-2}}$.

$\begin{aligned}v &= \sqrt{\frac{2\, (\text{KE})}{m}} \\ &= \sqrt{\frac{2\, (2.7 \times 10^{3}\; {\rm kg\cdot m^{2}\cdot s^{-2}})}{(600\; {\rm kg})}} \\ &= 3.0\; {\rm m\cdot s^{-1}}\end{aligned}$.

In other words, the velocity of this truck would be $3.0\; {\rm m\cdot s^{-1}}$.