Question

A 3 kg rock sits on a 0.8 meter ledge. If it is pushed off, how fast will it be going at the bottom?

Answer 1

The rock will be going approximately 3.96 m/s when it reaches the bottom.

Step-by-step explanation:

When the rock is pushed off the ledge, it will fall due to gravity. The velocity of the rock can be calculated using the equation:

v = sqrt(2 imes g imes h)

Where v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the ledge. Plugging in the values, we get:

v = sqrt(2 imes 9.8 imes 0.8) = sqrt(15.68) ≈ 3.96 m/s

Therefore, the rock will be going approximately 3.96 m/s when it reaches the bottom.

Answer 2

As long as it sits on the shelf, its potential energy
relative to the floor is . . .

Potential energy = (mass) x (gravity) x (height) =

(3 kg) x (9.8 m/s²) x (0.8m) = 23.52 joules .

If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.

Kinetic energy = (1/2) x (mass) x (speed)²

23.52 joules = (1/2) x (3 kg) x (speed)²

Divide each side by 1.5 kg : 23.52 m²/s² = speed²

Take the square root of each side: speed = √(23.52 m²/s²) = 4.85 m/s (rounded)