Question

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.

Answer 1

Answer:

Part a)

Final speed of the corn is 19.05 m/s

Part b)

Kinetic energy of the corn is 3.1 J

Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

now we also know that it will fall from rest and moving under constant acceleration so we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.81)(18.5)$

$v_f = 19.05 m/s$

Part b)

Kinetic energy of the corn is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.017)(19.05)^2$

$K = 3.1 J$