Question

A 25 Kg rock is sitting on a cliff that is 50 meters above the ground. If it fell off of the cliff, and all energy was conserved, what would be the
velocity of the rock right before it hit the ground?

Answer 1

Answer:

v = 31.32 [m/s]  

Explanation:

To solve this problem we must use the principle of energy conservation, which tells us that potential energy is converted into kinetic energy or vice versa. The potential energy can be calculated by the product of mass by gravity by height.

$E_{pot}=m*g*h$

where:

Epot = potential energy [J]

m = mass = 25 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 50 [m]

Now replacing:

$E_{pot}=25*9.81*50\\E_{pot}= 12262.5[J]$

When the rock falls the potential energy is converted into kinetic energy.

$E_{pot}=E_{k}\\E_{k}=\frac{1}{2}*m*v^{2}$

where:

Ek = kinetic energy [J]

v = velocity [m/s]

Now clearing v:

$v^{2} =\frac{E_{k}*2}{m}\\v=\sqrt{(2*12262)/25}\\v = 31.32 [m/s]$