What is the kinetic energy of an object with a mass of 3 kg that has been in freefall for  7 s?

Feb 26, 2017

7kJ of energy

Explanation:

The acceleration on the object is g, or 9.8m/s (assuming no air resistance) This number is a constant that applies to all objects close to the Earth.

Use the equation:

${v}_{f} = {v}_{i} + a t$ (The final velocity is the initial velocity plus the acceleration times time)

In this case, the initial velocity is zero (The object is dropped, not thrown downward) So:

${v}_{f} = a t$

${v}_{f} = \text{gt}$

${v}_{f} = 9.8 \left(\frac{m}{s} ^ 2\right) \cdot 7 s$

${v}_{f} = 68.6 \frac{m}{s}$

Now for kinetic energy = $\left(\frac{1}{2} m {v}^{2}\right)$

$K E = \frac{1}{2} \cdot 3 \text{kg} \cdot {\left(68.6 \left(\frac{m}{s}\right)\right)}^{2}$

Which is about $7059$ or $7000 k J$