a 15kg rock falls from the top of a building 8.0m high. Find its PE and KE
a.when at rest at the top of the building
b. halfway down the ground
c. when it reaches the ground
d.speed when it reaches the ground
make it clear for the answer
use 9.8 only
At time t seconds, the speed and height are given by
v = -9.8t
h = 8.0 - 4.9t^2
Then use the fact the the PE+KE must be constant
mgh + 1/2 mv^2 must stay the same.
To find the potential energy (PE) and kinetic energy (KE) at different points, we can use the formulas:
PE = mgh
KE = 0.5mv^2
where:
m = mass of the object (15 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height above the reference point (in this case, the ground)
v = velocity of the object
Let's calculate the values for each case:
a. When the rock is at rest at the top of the building (h = 8.0 m):
PE = mgh = 15 kg * 9.8 m/s^2 * 8.0 m = 1176 J
KE = 0
b. When the rock is halfway down the building (h = 8.0 m / 2 = 4.0 m):
PE = mgh = 15 kg * 9.8 m/s^2 * 4.0 m = 588 J
KE = 0
c. When the rock reaches the ground (h = 0 m):
PE = mgh = 15 kg * 9.8 m/s^2 * 0 m = 0 J
KE = 0.5mv^2
d. To find the speed of the rock when it reaches the ground, we need to use the conservation of energy principle. At the top of the building, all the potential energy is converted into kinetic energy when it reaches the ground. Therefore:
PE (at the top) = KE (at the bottom)
mgh = 0.5mv^2
Cancelling the mass (m) on both sides and solving for v:
gh = 0.5v^2
v^2 = 2gh
v = √(2gh)
Substituting the values:
v = √(2 * 9.8 m/s^2 * 8.0 m) = √(156.8) ≈ 12.5 m/s
Therefore, the speed of the rock when it reaches the ground is approximately 12.5 m/s.