A 500kg pig is standing at the top of a muddy hill on a rainy day. the hill is 100.0m long with a vertical drop of 30.0m. the pig slips and begins to slidedown the hill. What is the pig's speed at the bottom of the hill? Use the Law of conservation of Energy.
To determine the pig's speed at the bottom of the hill using the Law of Conservation of Energy, we need to consider the change in potential energy and kinetic energy.
The potential energy (PE) at the top of the hill is given by:
PE = mgh
where m is the mass of the pig (500 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical drop (30.0 m).
PE = (500 kg)(9.8 m/s^2)(30.0 m)
PE = 147,000 J
According to the Law of Conservation of Energy, this potential energy will be converted into kinetic energy (KE) at the bottom of the hill.
KE = (1/2)mv^2
where v is the speed of the pig at the bottom of the hill.
KE = (1/2)(500 kg)(v^2)
KE = 250v^2
Since energy is conserved, we can equate the potential energy at the top to the kinetic energy at the bottom:
PE = KE
147,000 J = 250v^2
Now we can solve for the velocity (v):
147,000 J = 250v^2
v^2 = 147,000 J / 250
v^2 = 588 m^2/s^2
Taking the square root of both sides, we find:
v = √(588 m^2/s^2)
v ≈ 24.24 m/s
Therefore, the pig's speed at the bottom of the hill is approximately 24.24 m/s.
To find the pig's speed at the bottom of the hill, we can use the Law of Conservation of Energy. According to this law, the total mechanical energy of an object (in this case, the pig) remains constant if only conservative forces are acting on it. In this scenario, we can assume that the only conservative force acting on the pig is gravitational potential energy.
First, let's calculate the potential energy of the pig at the top of the hill using the formula:
Potential energy = mass * gravity * height
Given:
Mass of the pig (m) = 500 kg
Gravity (g) = 9.8 m/s²
Height of the hill (h) = 30.0 m
Potential energy at the top = 500 kg * 9.8 m/s² * 30.0 m = 147000 J (joules)
Since the total mechanical energy at the top of the hill is equal to the total mechanical energy at the bottom (due to the conservation of energy), we can equate the potential energy at the top to the kinetic energy at the bottom of the hill.
Kinetic energy = 1/2 * mass * velocity²
We know that the mass of the pig (m) = 500 kg, and the height of the hill (h) is converted into kinetic energy at the bottom.
So, potential energy at the top = kinetic energy at the bottom
147000 J = 1/2 * 500 kg * velocity²
Now, we can solve for the velocity (v):
2 * 147000 J / 500 kg = velocity²
294 J/kg = velocity²
Taking the square root of both sides:
√(294 J/kg) = velocity
Therefore, the pig's speed at the bottom of the hill is approximately 17.13 m/s.
Mechanical energy would not be cnserved in such a situation. You have to consider the frictional force associated with sliding in mud.
However, if you want the book's expected answer, use
V = sqrt(2gH)
where H is the height of the hill.
The pig's mass and the length of the slope do not matter.