# What is the kinetic energy of an object with a mass of 6.25xx10^4 "g" having a speed of "26.3 m/s"?

Jul 2, 2015

The person's kinetic energy is equal to 21.6 kJ.

#### Explanation:

Kinetic energy is simply a measure of how much work needs to be done in order to accelerate an object from complete rest to a given velocity.

Mathematically, kinetic energy is defined as

${K}_{E} = \frac{1}{2} \cdot m \cdot {v}^{2}$, where

$m$ - the mass of the object - expressed in kilograms;
$v$ - the velocity of the object - expressed in meters per second.

So, convert the mass of the person from grams to kilograms

6.25 * 10^4cancel("g") * "1 kg"/(10^3cancel("g")) = "62.5 kg"

Now plug your values into the equation and solve for ${K}_{E}$

${K}_{E} = \frac{1}{2} \cdot {\text{62.5 kg" * 26.2""^2 "m"^2/"s}}^{2}$

K_E = 1/2 * 62.5 * 26.2""^2 * underbrace(("kg" * "m"^2)/"s"^2)_(color(blue)("=Joule"))

${K}_{E} = \text{21615. 3 J}$

Rounded to three sig figs and expressed in kilojoules, the answer will be

${K}_{E} = 21615.3 \cancel{\text{J") * "1 kJ"/(1000cancel("J")) = color(green)("21.6 kJ}}$

Jul 3, 2015

$\text{21600 J}$ (Rounded to three significant figures.)

#### Explanation:

$\text{KE} = \frac{1}{2} \cdot m \cdot {v}^{2}$, where KE is kinetic energy in Joules, $m$ is mass in kilograms, and ${v}^{2}$ is speed in m/s squared.

Known

mass = (6.25xx10^4cancel"g")xx(1"kg")/(1000cancel"g")="62.5 kg"
speed = $\text{26.3 m/s}$

Unknown
Kinetic energy

Solution
$\text{KE} = \frac{1}{2} \cdot m \cdot {v}^{2}$

"KE"=1/2*62.5"kg"*(26.3"m/s")^2

$\text{KE}$$= 21600 \text{kg·m"^2"/s"^2}$=$21600 \text{J}$
(Rounded to three significant figures.)