Answer to Question #87355 in Mechanics | Relativity for grace

Question

A 5.38 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the final kinetic energy (in J) for the first meter of the fall.

Answer 1

Let's first find the velocity of the object at the end of the first meter of the fall from the kinematic equation:

$$v_f^2 = v_0^2 + 2gs,$$

here, $v_0 = 0$ is the initial velocity of the object (at rest), $v_f$ is the velocity of the object at the end of the first meter of the fall, $g$ is the acceleration due to gravity and $s$ is the distance traveled by the object.

Then, we get:

$$v_f = \sqrt{2gs} = \sqrt{2 \cdot 9.8 \dfrac{m}{s^2} \cdot 1 m} = 4.43 \dfrac{m}{s}.$$

Finally, from the definition of the kinetic energy, we can calculate the final kinetic energy for the first meter of the fall:

$$KE_f = \dfrac{1}{2}mv_f^2 = \dfrac{1}{2} \cdot 5.38 kg \cdot (4.43 \dfrac{m}{s})^2 = 52.8 J.$$

Answer:

$KE_f = 52.8 J.$