| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the sled's speed when the spring returns to its uncompressed length, we can use the conservation of energy principle. The potential energy stored in the compressed spring is converted into kinetic energy of the sled as it moves. The potential energy (PE) stored in the spring when it is compressed by a distance x is given by the formula: \[ PE = \frac{1}{2} k x^2 \] where k is the spring constant and x is the compression distance. Given: k = 40.0 N/cm = 4000 N/m (since 1 N/cm = 100 N/m) x = 0.375 m Substituting the given values into the formula, we get: \[ PE = \frac{1}{2} \times 4000 \times (0.375)^2 = 281.25 \, \text{J} \] This potential energy is converted into kinetic energy (KE) of the sled when the spring returns to its uncompressed length. The kinetic energy of the sled is given by: \[ KE = \frac{1}{2} m v^2 \] where m is the mass of the sled and v is its velocity. Since the total energy is conserved, we can equate the potential energy to the kinetic energy: \[ PE = KE \] \[ 281.25 = \frac{1}{2} \times 70 \times v^2 \] Solving for v: \[ v^2 = \frac{2 \times 281.25}{70} = 8.0357 \] \[ v = \sqrt{8.0357} = 2.834 \, \text{m/s} \] Therefore, the sled's speed when the spring returns to its uncompressed length is approximately 2.834 m/s. |
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| deepseekr1 | To find the sled's speed when the spring returns to its uncompressed length, we use the principle of conservation of energy. The elastic potential energy stored in the compressed spring is converted entirely into the kinetic energy of the sled. 1. **Convert the spring constant to N/m**: \[ k = 40.0 \, \text{N/cm} = 40.0 \times 100 \, \text{N/m} = 4000 \, \text{N/m} \] 2. **Calculate the elastic potential energy (PE) stored in the spring**: \[ PE = \frac{1}{2} k x^2 \] where \( x = 0.375 \, \text{m} \): \[ PE = \frac{1}{2} \times 4000 \, \text{N/m} \times (0.375 \, \text{m})^2 \] \[ (0.375 \, \text{m})^2 = 0.140625 \, \text{m}^2 \] \[ PE = \frac{1}{2} \times 4000 \times 0.140625 = 2000 \times 0.140625 = 281.25 \, \text{J} \] 3. **Set the potential energy equal to the kinetic energy (KE)**: \[ KE = \frac{1}{2} m v^2 \] Solving for \( v \): \[ 281.25 \, \text{J} = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] \[ v^2 = \frac{2 \times 281.25}{70.0} = \frac{562.5}{70.0} \approx 8.0357 \] \[ v = \sqrt{8.0357} \approx 2.8368 \, \text{m/s} \] 4. **Round to three significant figures**: \[ v \approx 2.84 \, \text{m/s} \] The sled's speed when the spring returns to its uncompressed length is \(\boxed{2.84}\) m/s. |
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| gpto1 | To solve this problem, we'll use the principle of conservation of energy. Since the surface is frictionless and there's no air resistance, the mechanical energy of the system is conserved. This means the potential energy stored in the compressed spring will be entirely converted into the kinetic energy of the sled once the spring returns to its uncompressed length. **Step 1: Convert the Spring Constant to SI Units** The spring constant \( k \) is given in N/cm. We need to convert this to N/m to use SI units. \[ k = 40.0 \, \text{N/cm} \times 100 \, \frac{\text{cm}}{\text{m}} = 4000 \, \text{N/m} \] **Step 2: Write Down the Conservation of Energy Equation** The potential energy stored in the compressed spring is: \[ U_{\text{spring}} = \frac{1}{2} k x^2 \] where: - \( x \) is the compression distance. The kinetic energy of the sled after the spring is released is: \[ K_{\text{sled}} = \frac{1}{2} m v^2 \] where: - \( m \) is the mass of the sled and rider. - \( v \) is the final velocity of the sled. Setting the potential energy equal to the kinetic energy: \[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \] **Step 3: Solve for the Final Velocity \( v \)** Simplify the equation: \[ k x^2 = m v^2 \] Now solve for \( v \): \[ v = \sqrt{\frac{k x^2}{m}} \] **Step 4: Plug in the Known Values** Given: - \( k = 4000 \, \text{N/m} \) - \( x = 0.375 \, \text{m} \) - \( m = 70.0 \, \text{kg} \) Calculate \( x^2 \): \[ x^2 = (0.375\, \text{m})^2 = 0.140625\, \text{m}^2 \] Now plug the values into the equation for \( v \): \[ v = \sqrt{\frac{4000\, \text{N/m} \times 0.140625\, \text{m}^2}{70.0\, \text{kg}}} \] **Step 5: Perform the Calculations** Calculate the numerator: \[ k x^2 = 4000\, \text{N/m} \times 0.140625\, \text{m}^2 = 562.5\, \text{N}\cdot \text{m} \] Compute \( v^2 \): \[ v^2 = \frac{562.5\, \text{N}\cdot \text{m}}{70.0\, \text{kg}} = 8.03571429\, \text{m}^2/\text{s}^2 \] Calculate \( v \): \[ v = \sqrt{8.03571429\, \text{m}^2/\text{s}^2} \approx 2.836\, \text{m/s} \] **Answer:** Approximately **2.84 meters per second** is the sled's speed when the spring returns to its uncompressed length. |
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| GPT-4o | To determine the sled's speed when the spring returns to its uncompressed length, we can use the principles of energy conservation. Specifically, the potential energy stored in the compressed spring will be converted into the kinetic energy of the sled and rider. ### Step-by-Step Solution: 1. **Identify the given values:** - Spring constant, \( k = 40.0 \text{ N/cm} = 4000 \text{ N/m} \) (conversion from N/cm to N/m) - Compression of the spring, \( x = 0.375 \text{ m} \) - Mass of sled and rider, \( m = 70.0 \text{ kg} \) 2. **Calculate the potential energy stored in the compressed spring:** The potential energy (\( U \)) stored in a compressed or stretched spring is given by Hooke's Law: \[ U = \frac{1}{2} k x^2 \] Substituting the given values: \[ U = \frac{1}{2} \times 4000 \text{ N/m} \times (0.375 \text{ m})^2 \] 3. **Perform the calculation:** \[ U = \frac{1}{2} \times 4000 \times 0.140625 \] \[ U = 2000 \times 0.140625 \] \[ U = 281.25 \text{ J} \] 4. **Convert the potential energy to kinetic energy:** When the spring returns to its uncompressed length, all the potential energy will be converted into kinetic energy (\( K \)) of the sled and rider: \[ K = \frac{1}{2} m v^2 \] Since \( K = U \): \[ \frac{1}{2} m v^2 = 281.25 \text{ J} \] 5. **Solve for the velocity \( v \):** \[ \frac{1}{2} \times 70.0 \text{ kg} \times v^2 = 281.25 \text{ J} \] \[ 35.0 \text{ kg} \times v^2 = 281.25 \text{ J} \] \[ v^2 = \frac{281.25 \text{ J}}{35.0 \text{ kg}} \] \[ v^2 = 8.0357 \text{ m}^2/\text{s}^2 \] \[ v = \sqrt{8.0357 \text{ m}^2/\text{s}^2} \] \[ v \approx 2.83 \text{ m/s} \] ### Conclusion: The sled's speed when the spring returns to its uncompressed length is approximately \( 2.83 \text{ m/s} \). |
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| llama3.1-70B-Instruct | A fun problem! Let's dive in. We can solve this problem using the concept of conservation of energy. The energy stored in the compressed spring is converted into kinetic energy of the sled and rider as the spring returns to its uncompressed length. The energy stored in a compressed spring is given by: U = (1/2)kx^2 where k is the force constant, and x is the compression distance. In this case, k = 40.0 N/cm = 4000 N/m (converting to meters), and x = 0.375 m. U = (1/2) × 4000 N/m × (0.375 m)^2 = 281.25 J This energy is converted into kinetic energy of the sled and rider as the spring returns to its uncompressed length. The kinetic energy of an object is given by: K = (1/2)mv^2 where m is the mass of the object, and v is its velocity. We know the mass of the sled and rider: m = 70.0 kg. We want to find the velocity v when the spring returns to its uncompressed length, so we set the kinetic energy equal to the energy stored in the spring: (1/2)mv^2 = U Substituting the values, we get: (1/2) × 70.0 kg × v^2 = 281.25 J Simplifying and solving for v, we get: v^2 = 2 × 281.25 J / 70.0 kg = 8.036 m^2/s^2 v = √(8.036 m^2/s^2) = 2.83 m/s Therefore, the sled's speed when the spring returns to its uncompressed length is approximately 2.83 m/s. |
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| llama3.1-405B-Instruct | To find the speed of the sled when the spring returns to its uncompressed length, we can use the principle of conservation of energy. The energy stored in the compressed spring is converted into kinetic energy of the sled and rider. First, let's calculate the energy stored in the compressed spring: Energy stored in the spring = (1/2)kx^2 where k is the force constant of the spring (40.0 N/cm = 4000 N/m) and x is the compression distance (0.375 m). Energy stored in the spring = (1/2)(4000 N/m)(0.375 m)^2 = 281.25 J Now, we can set up the equation for conservation of energy: Energy stored in the spring = Kinetic energy of the sled and rider 281.25 J = (1/2)mv^2 where m is the mass of the sled and rider (70.0 kg) and v is the speed of the sled when the spring returns to its uncompressed length. Solving for v, we get: v = sqrt(2 \* 281.25 J / 70.0 kg) = 2.83 m/s Therefore, the speed of the sled when the spring returns to its uncompressed length is 2.83 m/s. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to use the principle of conservation of mechanical energy. The initial energy stored in the compressed spring will be converted into the kinetic energy of the sled when the spring returns to its uncompressed length. Here are the steps to solve the problem: 1. **Determine the spring potential energy (PE_spring):** The potential energy stored in a compressed spring is given by the formula: \[ PE_{spring} = \frac{1}{2} k x^2 \] where \( k \) is the spring constant and \( x \) is the compression distance. Given: - \( k = 40.0 \, \text{N/cm} \) - \( x = 0.375 \, \text{m} \) First, convert the spring constant from N/cm to N/m: \[ k = 40.0 \, \text{N/cm} \times 100 \, \text{cm/m} = 4000 \, \text{N/m} \] Now, calculate the spring potential energy: \[ PE_{spring} = \frac{1}{2} \times 4000 \, \text{N/m} \times (0.375 \, \text{m})^2 \] \[ PE_{spring} = \frac{1}{2} \times 4000 \times 0.140625 \, \text{J} \] \[ PE_{spring} = 281.25 \, \text{J} \] 2. **Set up the equation for kinetic energy (KE_sled):** The kinetic energy of the sled when the spring returns to its uncompressed length is given by: \[ KE_{sled} = \frac{1}{2} m v^2 \] where \( m \) is the mass of the sled and rider, and \( v \) is the speed of the sled. Given: - \( m = 70.0 \, \text{kg} \) According to the conservation of mechanical energy, the initial spring potential energy is equal to the final kinetic energy of the sled: \[ PE_{spring} = KE_{sled} \] \[ 281.25 \, \text{J} = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] 3. **Solve for the speed \( v \):** \[ 281.25 = \frac{1}{2} \times 70.0 \times v^2 \] \[ 281.25 = 35.0 \times v^2 \] \[ v^2 = \frac{281.25}{35.0} \] \[ v^2 = 8.0357 \] \[ v = \sqrt{8.0357} \] \[ v \approx 2.834 \, \text{m/s} \] Therefore, the sled's speed when the spring returns to its uncompressed length is approximately \( 2.834 \, \text{m/s} \). |
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| Qwen2-72B-Instruct | To find the sled's speed when the spring returns to its uncompressed length, we can use the conservation of energy principle. The potential energy stored in the compressed spring is converted into kinetic energy of the sled as it moves. The potential energy (PE) stored in the spring when it is compressed by a distance x is given by the formula: \[ PE = \frac{1}{2} k x^2 \] where k is the spring constant and x is the compression distance. Given: k = 40.0 N/cm = 4000 N/m (since 1 N/cm = 100 N/m) x = 0.375 m Substituting the given values into the formula, we get: \[ PE = \frac{1}{2} \times 4000 \times (0.375)^2 = 281.25 \, \text{J} \] This potential energy is converted into kinetic energy (KE) of the sled when the spring returns to its uncompressed length. The kinetic energy of the sled is given by: \[ KE = \frac{1}{2} m v^2 \] where m is the mass of the sled and v is its velocity. Since the total energy is conserved, we can equate the potential energy to the kinetic energy: \[ PE = KE \] \[ 281.25 = \frac{1}{2} \times 70 \times v^2 \] Solving for v: \[ v^2 = \frac{2 \times 281.25}{70} = 8.0357 \] \[ v = \sqrt{8.0357} = 2.834 \, \text{m/s} \] Therefore, the sled's speed when the spring returns to its uncompressed length is approximately 2.834 m/s. |
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| baichuan4 | To solve this problem, we can use the principle of conservation of energy. When the spring is compressed, it stores potential energy. When it expands back to its original length, this potential energy is converted into kinetic energy of the sled. The potential energy stored in the spring is given by the equation: \[ U_s = \frac{1}{2} k x^2 \] where \( U_s \) is the potential energy of the spring, \( k \) is the spring constant, and \( x \) is the amount of compression of the spring. The kinetic energy of the sled is given by the equation: \[ K = \frac{1}{2} m v^2 \] where \( K \) is the kinetic energy of the sled, \( m \) is the mass of the sled and rider, and \( v \) is the speed of the sled. Since the total energy is conserved, we can set the initial potential energy of the spring equal to the final kinetic energy of the sled: \[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \] Now, we can substitute the given values: \[ \frac{1}{2} \times 40.0 \, \text{N/cm} \times (0.375 \, \text{m})^2 = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] Note that we need to convert the spring constant from N/cm to N/m by multiplying by 100, since the compression distance is given in meters: \[ \frac{1}{2} \times 4000 \, \text{N/m} \times (0.375)^2 = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] Calculating the left-hand side: \[ 4000 \times 0.375^2 = 562.5 \, \text{J} \] Now, we can solve for \( v \): \[ 562.5 = \frac{1}{2} \times 70.0 \times v^2 \] \[ v^2 = \frac{562.5 \times 2}{70.0} \] \[ v^2 = 16.83 \, \text{m}^2/\text{s}^2 \] \[ v = \sqrt{16.83} \] \[ v = 4.1 \, \text{m/s} \] So, the sled's speed when the spring returns to its uncompressed length is approximately 4.1 m/s. |
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| doubao | The potential energy stored in the spring when it is compressed is given by 1/2 kx^2, where k is the force constant and x is the compression distance. Converting the force constant to SI units: k = 40.0 N/cm = 4000 N/m. The potential energy stored = 1/2 * 4000 * (0.375)^2 = 281.25 J. When the spring returns to its uncompressed length, this potential energy is converted to the kinetic energy of the sled and rider. The kinetic energy is given by 1/2 mv^2, where m is the mass and v is the velocity. So 1/2 * 70 * v^2 = 281.25. 70v^2 = 562.5. v^2 = 8.036. v = 2.83 m/s. So the sled's speed when the spring returns to its uncompressed length is 2.83 m/s. |
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| minimax | To solve this problem, we can use the principle of conservation of energy. The potential energy stored in the compressed spring will be converted into kinetic energy of the sled when the spring is released. First, we need to convert the force constant from N/cm to N/m, since the standard unit for force constant is Newtons per meter (N/m). Force constant k = 40.0 N/cm = 40.0 N / 0.01 m = 4000 N/m The potential energy (PE) stored in a compressed spring is given by the formula: PE = 1/2 * k * x^2 where k is the force constant and x is the compression distance. Let's calculate the potential energy: PE = 1/2 * 4000 N/m * (0.375 m)^2 PE = 1/2 * 4000 * 0.140625 PE = 2000 * 0.140625 PE = 281.25 J This potential energy will be converted into kinetic energy (KE) of the sled when the spring is uncompressed. The kinetic energy of an object is given by: KE = 1/2 * m * v^2 where m is the mass of the sled and rider, and v is the velocity we want to find. Since the potential energy will be equal to the kinetic energy at the point when the spring is uncompressed (assuming no energy loss to friction or air resistance), we can set PE equal to KE: PE = KE 281.25 J = 1/2 * 70.0 kg * v^2 Now we solve for v: 281.25 J = 35 kg * v^2 v^2 = 281.25 J / 35 kg v^2 = 8.*** m^2/s^2 Taking the square root of both sides to solve for v: v = sqrt(8.*** m^2/s^2) v ≈ 2.834 m/s Therefore, the sled's speed when the spring returns to its uncompressed length is approximately 2.834 m/s. |
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| kimi | To solve this problem, we can use the conservation of energy principle. The potential energy stored in the compressed spring will be converted into kinetic energy as the sled moves. The potential energy (PE) stored in the spring when it is compressed is given by the formula: \[ PE = \frac{1}{2} k x^2 \] where \( k \) is the spring constant and \( x \) is the distance the spring is compressed. Given: \( k = 40.0 \) N/cm = \( 40.0 \times 10^2 \) N/m (since 1 N/cm = \( 10^2 \) N/m) \( x = 0.375 \) m \( m = 70.0 \) kg First, we calculate the potential energy stored in the spring: \[ PE = \frac{1}{2} k x^2 \] \[ PE = \frac{1}{2} \times 40.0 \times 10^2 \times (0.375)^2 \] \[ PE = \frac{1}{2} \times 4000 \times 0.140625 \] \[ PE = 2000 \times 0.140625 \] \[ PE = 281.25 \] J (Joules) When the spring returns to its uncompressed length, all the potential energy will have been converted into kinetic energy (KE) of the sled. The kinetic energy is given by the formula: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is the mass of the sled and rider, and \( v \) is the velocity of the sled. Since the potential energy is equal to the kinetic energy at that point, we can set the two equal to each other and solve for \( v \): \[ KE = PE \] \[ \frac{1}{2} m v^2 = PE \] \[ \frac{1}{2} \times 70.0 \times v^2 = 281.25 \] \[ 35.0 \times v^2 = 281.25 \] \[ v^2 = \frac{281.25}{35.0} \] \[ v^2 = 8.0357 \] \[ v = \sqrt{8.0357} \] \[ v \approx 2.83 \] m/s So, the sled's speed when the spring returns to its uncompressed length is approximately \( 2.83 \) m/s. |
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| wenxinyiyan | To solve this problem, we will use conservation of energy. The potential energy stored in the compressed spring will be converted into kinetic energy as the sled and rider are pushed away. First, we need to calculate the potential energy stored in the spring. The spring force is given by Hooke's Law: F = kx where F is the force, k is the force constant, and x is the displacement. The potential energy stored in the spring is given by: PE = (1/2)kx^2 where PE is the potential energy, k is the force constant, and x is the displacement. Given: k = 40.0 N/cm = 4000 N/m (since 1 m = 100 cm) x = 0.375 m m = 70.0 kg Calculate the potential energy stored in the spring: PE = (1/2)(4000 N/m)(0.375 m)^2 PE = (1/2)(4000 N/m)(0.140625 m^2) PE = 281.25 J Now, we will use conservation of energy. The potential energy stored in the spring will be converted into kinetic energy as the sled and rider are pushed away. Kinetic energy is given by: KE = (1/2)mv^2 where KE is the kinetic energy, m is the mass, and v is the velocity. Set the potential energy equal to the kinetic energy: PE = KE 281.25 J = (1/2)(70.0 kg)v^2 Solve for v: v^2 = 281.25 J / (1/2)(70.0 kg) v^2 = 8.035714286 m^2/s^2 v = sqrt(8.035714286 m^2/s^2) v = 2.83 m/s Therefore, the sled's speed when the spring returns to its uncompressed length is 2.83 m/s. |
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| zhipu | To solve this problem, we can use the principle of conservation of energy. The potential energy stored in the compressed spring will be converted into the kinetic energy of the sled and rider as the spring returns to its uncompressed length. The potential energy (PE) stored in a spring is given by Hooke's Law: \[ PE = \frac{1}{2} k x^2 \] where: - \( k \) is the spring constant, - \( x \) is the compression of the spring. First, we need to convert the spring constant from N/cm to N/m to match the units of the compression distance: \[ k = 40.0 \, \text{N/cm} = 40.0 \, \text{N/cm} \times \frac{100 \, \text{cm}}{1 \, \text{m}} = 4000 \, \text{N/m} \] Now we can calculate the potential energy stored in the spring when it is compressed by 0.375 m: \[ PE = \frac{1}{2} \times 4000 \, \text{N/m} \times (0.375 \, \text{m})^2 \] \[ PE = \frac{1}{2} \times 4000 \times 0.140625 \] \[ PE = 2000 \times 0.140625 \] \[ PE = 281.25 \, \text{J} \] This potential energy will be equal to the kinetic energy (KE) of the sled and rider when the spring is no longer compressed: \[ KE = \frac{1}{2} m v^2 \] where: - \( m \) is the mass of the sled and rider, - \( v \) is the velocity of the sled and rider. Setting the potential energy equal to the kinetic energy, we get: \[ 281.25 \, \text{J} = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] Now we solve for \( v \): \[ v^2 = \frac{2 \times 281.25 \, \text{J}}{70.0 \, \text{kg}} \] \[ v^2 = \frac{562.5}{70.0} \] \[ v^2 = 8.0357 \, \text{m}^2/\text{s}^2 \] Taking the square root of both sides to find \( v \): \[ v = \sqrt{8.0357} \] \[ v \approx 2.83 \, \text{m/s} \] So, the sled's speed when the spring returns to its uncompressed length is approximately \( 2.83 \, \text{m/s} \). |
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| gpto1 | To solve this problem, we'll use the principle of conservation of energy. Since the surface is frictionless and there's no air resistance, the mechanical energy of the system is conserved. This means the potential energy stored in the compressed spring will be entirely converted into the kinetic energy of the sled once the spring returns to its uncompressed length. **Step 1: Convert the Spring Constant to SI Units** The spring constant \( k \) is given in N/cm. We need to convert this to N/m to use SI units. \[ k = 40.0 \, \text{N/cm} \times 100 \, \frac{\text{cm}}{\text{m}} = 4000 \, \text{N/m} \] **Step 2: Write Down the Conservation of Energy Equation** The potential energy stored in the compressed spring is: \[ U_{\text{spring}} = \frac{1}{2} k x^2 \] where: - \( x \) is the compression distance. The kinetic energy of the sled after the spring is released is: \[ K_{\text{sled}} = \frac{1}{2} m v^2 \] where: - \( m \) is the mass of the sled and rider. - \( v \) is the final velocity of the sled. Setting the potential energy equal to the kinetic energy: \[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \] **Step 3: Solve for the Final Velocity \( v \)** Simplify the equation: \[ k x^2 = m v^2 \] Now solve for \( v \): \[ v = \sqrt{\frac{k x^2}{m}} \] **Step 4: Plug in the Known Values** Given: - \( k = 4000 \, \text{N/m} \) - \( x = 0.375 \, \text{m} \) - \( m = 70.0 \, \text{kg} \) Calculate \( x^2 \): \[ x^2 = (0.375\, \text{m})^2 = 0.140625\, \text{m}^2 \] Now plug the values into the equation for \( v \): \[ v = \sqrt{\frac{4000\, \text{N/m} \times 0.140625\, \text{m}^2}{70.0\, \text{kg}}} \] **Step 5: Perform the Calculations** Calculate the numerator: \[ k x^2 = 4000\, \text{N/m} \times 0.140625\, \text{m}^2 = 562.5\, \text{N}\cdot \text{m} \] Compute \( v^2 \): \[ v^2 = \frac{562.5\, \text{N}\cdot \text{m}}{70.0\, \text{kg}} = 8.03571429\, \text{m}^2/\text{s}^2 \] Calculate \( v \): \[ v = \sqrt{8.03571429\, \text{m}^2/\text{s}^2} \approx 2.836\, \text{m/s} \] **Answer:** Approximately **2.84 meters per second** is the sled's speed when the spring returns to its uncompressed length. |
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| deepseekv3 | To solve this problem, we'll use the principles of energy conservation. The potential energy stored in the compressed spring is converted into the kinetic energy of the sled and rider when the spring returns to its uncompressed length. ### Given: - Spring force constant, \( k = 40.0 \, \text{N/cm} = 4000 \, \text{N/m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \)) - Mass of the sled and rider, \( m = 70.0 \, \text{kg} \) - Compression of the spring, \( x = 0.375 \, \text{m} \) ### Step 1: Calculate the Potential Energy Stored in the Spring The potential energy \( U \) stored in a compressed spring is given by: \[ U = \frac{1}{2} k x^2 \] Substitute the given values: \[ U = \frac{1}{2} \times 4000 \, \text{N/m} \times (0.375 \, \text{m})^2 \] \[ U = \frac{1}{2} \times 4000 \times 0.140625 \] \[ U = 2000 \times 0.140625 \] \[ U = 281.25 \, \text{J} \] ### Step 2: Convert Potential Energy into Kinetic Energy When the spring returns to its uncompressed length, all the potential energy is converted into kinetic energy \( K \) of the sled and rider: \[ K = \frac{1}{2} m v^2 \] Set the potential energy equal to the kinetic energy: \[ 281.25 \, \text{J} = \frac{1}{2} \times 70.0 \, \text{kg} \times v^2 \] ### Step 3: Solve for the Speed \( v \) Rearrange the equation to solve for \( v \): \[ v^2 = \frac{2 \times 281.25}{70.0} \] \[ v^2 = \frac{562.5}{70.0} \] \[ v^2 = 8.0357 \] \[ v = \sqrt{8.0357} \] \[ v \approx 2.84 \, \text{m/s} \] ### Final Answer: The sled's speed when the spring returns to its uncompressed length is approximately **2.84 m/s**. |
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