Question

question 2
the force of gravity on a 2 - kg rock is twice as great as that on a 1 - kg rock. if both are released from rest, why doesnt the heavier rock fall downwards faster? explain in terms of physics concepts/principles. (you do not need a separate written work submission for this question but instead can just type in your explanation here.)

Explanation:

According to Newton's second - law $F = ma$, where $F$ is the force, $m$ is the mass and $a$ is the acceleration. The gravitational force on an object is $F=mg$ (where $g$ is the acceleration due to gravity). For the 2 - kg rock, $F_1 = 2mg$ and for the 1 - kg rock, $F_2=mg$. When we calculate the acceleration $a=\frac{F}{m}$, for the 2 - kg rock $a_1=\frac{2mg}{2m}=g$, and for the 1 - kg rock $a_2=\frac{mg}{m}=g$. So both have the same acceleration $g$ and fall at the same rate.

Answer:

Both rocks have the same acceleration due to gravity. According to Newton's second - law $F = ma$, when calculating the acceleration of the falling objects considering the gravitational force $F = mg$, the mass cancels out, resulting in the same acceleration $g$ for both rocks, so they fall at the same rate.