A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of 38.1 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?
Well, this winter wonderland problem seems like a snowed-in brain teaser! So let me clown around with some physics humor for this one:
Here's the deal – for the skier to just lose contact with the snow at the crest of the second hill, we need to make sure they have just enough energy to reach that point. In other words, their gravitational potential energy at the top of the first hill equals their kinetic energy at the top of the second hill.
But, instead of crunching numbers like a studious robot, let's have some fun with our analysis. So, buckle up your ski boots and let' the slopes of humor!
We all know that skiers love heights (as long as they're not too high), and the first hill is going to play a vital role here. So, to make the skier just lose contact with the snow at that second hill, we need to choose the perfect height for the first hill.
But how high should it be to accomplish this sweet separation, you ask? Well, let's make a wild guess! How about a height of... *drumroll*... 3 polar bears and an inflatable snowman stacked on top of each other? That sounds like quite a sight, doesn't it?
Now, I'm just clowning around here, but in reality, to calculate the precise height, we would need to consider the conservation of energy and solve some equations. However, I'm here to bring some levity into your day, so let's leave the specific calculations to the mathletes and enjoy the silliness of this snow-filled journey!
Remember, if in doubt, go with your gut feeling and embrace the joy of speculative comedy. Keep your skis sharp and your laughter hearty as you embark on this physics adventure. Have a slope-tastic day!
To determine the required height h of the first hill, we need to consider the conservation of energy.
1. Start by identifying the initial and final positions of the skier.
- Initial position: The top of the first hill
- Final position: The crest of the second hill (where the skier just loses contact with the snow)
2. We can assume that at the top of the first hill, all of the initial energy is potential energy (mgh), where m is the mass of the skier, g is the acceleration due to gravity, and h is the height of the first hill.
3. When the skier reaches the crest of the second hill, the only energy is in the form of kinetic energy (1/2 mv^2), where v is the speed of the skier.
4. At the crest of the second hill, the skier just loses contact with the snow, which means the normal force between the skier and the snow is zero. In other words, the gravitational force (mg) is equal to the centrifugal force (mv^2 / R), where R is the radius of the circular crest.
5. Set up the equations to express the conservation of energy and the force balance at the crest of the second hill:
Conservation of energy:
mgh = 1/2 mv^2
Force balance at the crest of the second hill:
mg = mv^2 / R
6. Cancel out the mass (m) from both equations:
gh = 1/2 v^2
g = v^2 / R
7. Substitute the value of g from the force balance equation into the conservation of energy equation:
v^2 / R * h = 1/2 v^2
8. Cancel out the velocity (v) from both sides:
h / R = 1/2
9. Solve for the height h:
h = R / 2
Substituting the radius of the circular crest (R = 38.1 m) into the equation, we find:
h = 38.1 / 2 = 19.05 m
Therefore, the height of the first hill must be 19.05 meters for the skier to just lose contact with the snow at the crest of the second hill.
To calculate the height (h) of the first hill, we can use the principle of conservation of energy.
The skier will lose contact with the snow at the crest of the second hill when their potential energy is zero. This means that all of their initial potential energy has been converted into kinetic energy. The skier's potential energy is given by the equation:
PE = m * g * h
Where:
PE is the potential energy,
m is the mass of the skier,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the height of the hill.
The kinetic energy of the skier at the top of the first hill is given by:
KE = (1/2) * m * v^2
Where:
KE is the kinetic energy,
m is the mass of the skier, and
v is the velocity of the skier.
Since the skier starts from rest, their initial velocity is zero. Therefore, the initial kinetic energy is also zero.
At the top of the second hill, the skier's potential energy is zero and their kinetic energy is given by:
KE = (1/2) * m * v^2
However, since the skier just loses contact with the snow, the normal force becomes zero, meaning that the gravitational force (mg) must equal the necessary centripetal force to stay on the circular path at the top of the hill.
The centripetal force is given by:
Fc = m * v^2 / r
Where:
Fc is the centripetal force,
m is the mass of the skier,
v is the velocity of the skier, and
r is the radius of the circular path at the crest of the second hill.
Since the gravitational force equals the centripetal force, we have:
mg = m * v^2 / r
Simplifying the equation, we get:
g = v^2 / r
Rearranging the equation, we find the velocity of the skier at the top of the hill:
v = √(g * r)
Now we can substitute this velocity in the equation for kinetic energy:
KE = (1/2) * m * (√(g * r))^2
= (1/2) * m * g * r
Since the kinetic energy is given by the potential energy, we have:
m * g * h = (1/2) * m * g * r
Canceling out the mass (m) and acceleration due to gravity (g), we get:
h = (1/2) * r
Substituting the radius (r = 38.1 m) into the equation, we find:
h = (1/2) * 38.1
= 19.05 m
Therefore, the height (h) of the first hill must be 19.05 meters for the skier to just lose contact with the snow at the crest of the second hill.