A 58.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.20 m. What speed does he need at the bottom?
at the top, PE = mgh = 58.0 * 9.81 * 2.20
so KE = 1/2 mv^2 must be at least that much at the bottom
Well, I'm no mathematician, but I'll give it a shot. If you want to calculate the speed the skateboarder needs at the bottom of the quarter pipe, you can use conservation of energy.
At the bottom, the skateboarder has both kinetic energy and gravitational potential energy. When he reaches the top, all of his kinetic energy is converted to potential energy, and vice versa at the bottom.
So, let's do some circus math. The gravitational potential energy at the top is equal to the gravitational potential energy at the bottom plus the change in potential energy. Since the top and bottom are at the same height, there is no change in potential energy, which means all of it is converted into kinetic energy at the bottom.
Using the formula for gravitational potential energy (mgh) and kinetic energy (1/2mv²), we can set them equal to each other:
mgh = 1/2mv²
Oops! Looks like I made a clownish error. I forgot to include the velocity of the skateboarder squaring around that quarter pipe. Let me fix that equation:
mgh = 1/2mv² + 1/2m(2πr)²
Now, we just plug in the given values and solve for the velocity (v):
58.0 kg * 9.8 m/s² * 2.20 m = 1/2 * 58.0 kg * v² + 1/2(58.0 kg)(2π × 2.20 m)²
After doing all the math, the final result is...
Sorry, I seem to have slipped on a banana peel there. As a clown bot, I specialize in humor, not advanced mathematics. But don't worry! I'm sure you can find a trusty physics textbook or a helpful online resource to guide you through the calculation. Good luck, my friend!
To find the speed the skateboarder needs at the bottom to reach the upper edge of the quarter pipe, we can use the law of conservation of energy.
The total energy at the bottom of the quarter pipe is given by the sum of the kinetic energy and potential energy:
E(total) = KE + PE
At the bottom of the quarter pipe, all the energy is in the form of kinetic energy because the potential energy is zero:
E(total) = KE
The kinetic energy is given by the equation:
KE = 1/2 * m * v^2
Where:
m = mass of the skateboarder = 58.0 kg
v = velocity or speed of the skateboarder
Since the quarter pipe is curved, the velocity needs to have both horizontal and vertical components. The vertical component of the velocity will help the skateboarder to stay on the track.
Using the conservation of energy, we can equate this equation with the potential energy equation at the top of the track:
PE(top) = m * g * h
Where:
PE(top) = potential energy at the top of the quarter pipe
m = mass of the skateboarder = 58.0 kg
g = acceleration due to gravity = 9.8 m/s^2 (approximately)
h = height of the quarter pipe
In this case, the height of the quarter pipe is equal to the radius of the quarter pipe because the upper edge of the quarter pipe is at the same height as the starting point:
h = 2.20 m
Now, we can set up the equation:
1/2 * m * v^2 = m * g * h
Simplifying and solving for v:
v^2 = 2 * g * h
v = sqrt(2 * g * h)
Substituting the values:
v = sqrt(2 * 9.8 m/s^2 * 2.20 m)
v ≈ sqrt(43.12) m/s
v ≈ 6.56 m/s
Therefore, the skateboarder needs a speed of approximately 6.56 m/s at the bottom to reach the upper edge of the quarter pipe.
To find the speed the skateboarder needs at the bottom of the quarter pipe, we can use the principle of conservation of energy.
At the bottom of the quarter pipe, the skateboarder will have both kinetic energy (due to motion) and potential energy (due to height). We can equate these two energies as follows:
Kinetic Energy at the bottom = Potential Energy at the top
The potential energy at the top of the quarter pipe is equal to the gravitational potential energy given by:
Potential Energy = mass * acceleration due to gravity * height
The height can be found by considering the quarter pipe as a quarter of a circle with a radius of 2.20 m. The height of the quarter pipe is the same as the radius, which is 2.20 m. We can plug these values into the potential energy equation:
Potential Energy = 58.0 kg * 9.8 m/s^2 * 2.20 m
Now, let's solve for the kinetic energy at the bottom. The kinetic energy is given by:
Kinetic Energy = (1/2) * mass * velocity^2
Since the skateboarder wants to just make it to the upper edge and not go above it, the potential energy at the top is zero. Therefore, the kinetic energy at the bottom is also zero. Hence, we can set the two equations equal to each other:
0 = 58.0 kg * 9.8 m/s^2 * 2.20 m + (1/2) * 58.0 kg * velocity^2
We can now solve for the velocity. Rearranging the equation:
(1/2) * 58.0 kg * velocity^2 = -58.0 kg * 9.8 m/s^2 * 2.20 m
Dividing both sides by (1/2) * 58.0 kg:
velocity^2 = -9.8 m/s^2 * 2.20 m * 2
Taking the square root of both sides to find the velocity:
velocity = sqrt(-9.8 m/s^2 * 2.20 m * 2)
Calculating this value will give you the required speed the skateboarder needs at the bottom of the quarter pipe.