1. A student lifts his 2.0-kg pet rock 2.8m straight up. He then lets it drop to the ground. Use the Law of Conservation of Energy to calculate how fast the rock will be moving .....
a. half way down
b. just before it hits the ground
2. A 65-kg girl is running with a speed of 2.5 m/s. How much kinetic energy does she have? She grabs on to a rope that is hanging from the ceiling, and swings from the end of the rope. How high off the ground will she swing?
3. How much work must be done to increase the speed of a 12-kg bicycle ridden by a 68-kg rider from 8.2 m/s to 12.7 m/s?
1. As a clown bot, I prefer funny calculations rather than scientific ones. So, let's add a little humor to these questions!
1a. When the rock is halfway down, it'll be so confused that it won't know whether it's falling or going up. It will freeze in mid-air, leaving you wondering what kind of magic pet rock you have!
1b. Just before the rock hits the ground, it will be moving at an unbelievable speed. It'll break the sound barrier and create a mini-sonic boom, startling everyone in the vicinity. Make sure to warn your neighbors about the impending rock-nado!
2. The girl has a kinetic energy of approximately "a whole lot" joules. She's so full of energy that she could power a whole amusement park with her enthusiasm alone. Remember, caution is essential when swinging from ropes - it's all fun and games until you accidentally swing into a wall!
As for how high she'll swing, it depends on how catchy the tune she sings while swinging. The higher the pitch, the higher she flies! Just make sure she doesn't swing all the way to the moon.
3. To increase the speed of the bicycle ridden by the rider, you'll need an enchanted bicycle pump that adds turbo boost. Unfortunately, such pumps are yet to be invented. So, we'll have to rely on plain old physics and calculations.
The work required to increase the speed can be calculated using the formula: work = ½ * mass * (final velocity^2 - initial velocity^2)
In this case, it'll be a considerable amount of work, equivalent to the number of times you have yelled at your screen when your internet connection is slow. So, let's hope the rider gets a burst of motivation and pedals like their life depends on it!
1. To calculate how fast the rock will be moving, we can use the Law of Conservation of Energy.
a. Halfway down:
- At the top of the motion, the rock has only potential energy, given by the equation PE = mgh, where m is the mass of the rock (2.0 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (2.8 m). So, PE = (2.0 kg)(9.8 m/s²)(2.8 m) = 54.88 J.
- At halfway down, the rock has lost half of its potential energy, which is now converted into kinetic energy. So, the kinetic energy (KE) is half of the potential energy. KE = 1/2 * 54.88 J = 27.44 J.
- To find the velocity, we can use the equation KE = 1/2 * mv², where m is the mass and v is the velocity. Rearranging the equation, we get v = √(2KE / m) = √(2 * 27.44 J / 2.0 kg) ≈ 4.16 m/s.
b. Just before it hits the ground:
- At the top of the motion, the rock has only potential energy (PE), given by PE = mgh, where m is the mass of the rock (2.0 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height (2.8 m). So, PE = (2.0 kg)(9.8 m/s²)(2.8 m) = 54.88 J.
- Just before it hits the ground, the potential energy has been fully converted into kinetic energy (KE). So, KE = 54.88 J.
- Using the same equation as in part a, KE = 1/2 * mv², we can find the velocity. Rearranging the equation, we get v = √(2KE / m) = √(2 * 54.88 J / 2.0 kg) ≈ 6.57 m/s.
2. To calculate the kinetic energy of the girl and the height she will swing to, note that the Law of Conservation of Energy applies here as well.
- The formula for kinetic energy (KE) is KE = 1/2 * mv², where m is the mass (65 kg) and v is the velocity (2.5 m/s).
- Plugging in the values, KE = 1/2 * (65 kg) * (2.5 m/s)² = 203.125 J.
- When the girl grabs onto the rope, the kinetic energy is converted into potential energy as she swings upward.
- Using the equation for potential energy (PE) PE = mgh, where m is the mass (65 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height she swings to. Rearranging the equation, we get h = PE / (mg).
- Substituting the values, h = 203.125 J / (65 kg * 9.8 m/s²) ≈ 0.324 m or 32.4 cm.
So, the girl will swing to a height of approximately 0.324 m or 32.4 cm off the ground.
3. To calculate the work required to increase the speed of the bicycle-rider system, we need to use the work-energy principle.
- The initial kinetic energy (KE₁) is given by KE₁ = 1/2 * m₁ * v₁², where m₁ is the mass of the bicycle (12 kg) and v₁ is the initial velocity (8.2 m/s).
- Plugging in the values, KE₁ = 1/2 * (12 kg) * (8.2 m/s)² = 401.76 J.
- The final kinetic energy (KE₂) is given by KE₂ = 1/2 * (m₁ + m₂) * v₂², where m₂ is the mass of the rider (68 kg) and v₂ is the final velocity (12.7 m/s).
- Plugging in the values, KE₂ = 1/2 * (12 kg + 68 kg) * (12.7 m/s)² = 6162.64 J.
- The work done (W) is given by W = KE₂ - KE₁ = 6162.64 J - 401.76 J = 5760.88 J.
So, the work required to increase the speed of the bicycle-rider system from 8.2 m/s to 12.7 m/s is approximately 5760.88 J.
To answer these questions, we can use the formulas and principles related to energy, work, and motion. Let's break down each question and its solution step by step:
1. For the first question, we'll use the Law of Conservation of Energy. This law states that the total energy of a system remains constant unless acted upon by an external force. In this case, we'll consider the potential energy of the rock being converted into kinetic energy as it falls.
a. To calculate the speed of the rock halfway down, we can use the conservation of energy principle. The potential energy at the initial position (lifting height) will be converted to kinetic energy at the halfway point.
- Potential energy: PE = mgh
Here, m = mass of the rock (2.0 kg), g = acceleration due to gravity (9.8 m/s²), and h = lifting height (2.8 m).
- The potential energy at the starting position (top) is PE = (2.0 kg)(9.8 m/s²)(2.8 m) = 54.88 J.
- At the halfway point, the potential energy will convert completely into kinetic energy, so the kinetic energy at this point will be KE = (1/2)mv².
- Rearranging the formula and substituting the known values, we get v = √(2KE/m).
- Plugging in the values, v = √[(2 × 54.88 J) / (2.0 kg)] = √(54.88 J / kg) = 4.17 m/s.
b. To calculate the speed just before the rock hits the ground, we'll use the same conservation of energy principle.
- At the starting position, the potential energy is the same as in part a, which is PE = 54.88 J.
- At the bottom, all the potential energy is converted into kinetic energy, so we have KE = (1/2)mv².
- Rearranging the equation and substituting the values, we get v = √(2KE/m).
- Plugging in the values, v = √[(2 × 54.88 J) / (2.0 kg)] = √(54.88 J / kg) = 4.17 m/s.
Therefore, the speed of the rock will be 4.17 m/s both halfway down and just before it hits the ground.
2. To calculate the kinetic energy and the height the girl will swing, we'll use the formulas for kinetic energy and conservation of mechanical energy.
- Kinetic energy: KE = (1/2)mv²
Here, m = mass of the girl (65 kg) and v = her speed (2.5 m/s).
Substituting these values, KE = (1/2)(65 kg)(2.5 m/s)² = 81.25 J.
- When she grabs the rope and swings, her kinetic energy will be converted into gravitational potential energy.
Gravitational potential energy: PE = mgh
Here, g = acceleration due to gravity (9.8 m/s²) and h = height off the ground.
So, PE = mgh = KE
Substituting the known values, 65 kg × 9.8 m/s² × h = 81.25 J.
Simplifying, h = 81.25 J / (65 kg × 9.8 m/s²) = 0.126 m.
Therefore, the girl will swing to a height of approximately 0.126 meters off the ground.
3. To calculate the work required to increase the speed of the bicycle, we'll use the work-energy theorem.
- Work: W = ΔKE
Here, ΔKE = change in kinetic energy.
Substituting the known values, ΔKE = (1/2)mv² - (1/2)mu²,
where m = mass of the bicycle and rider (12 kg + 68 kg), u = initial speed (8.2 m/s), and v = final speed (12.7 m/s).
Plugging in the values, ΔKE = (1/2)(80 kg)[(12.7 m/s)² - (8.2 m/s)²] = 2853 J.
Therefore, the work required to increase the speed of the bicycle from 8.2 m/s to 12.7 m/s is 2853 Joules.