▼ A 62 kg driver gets into an empty taptap to start the day's work. The springs compress 1.8x102 m. What is the effective spring constant of the system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures. ▸ View Available Hint(s) k= 3.4×104 N/m Submit Previous Answers Correct If you need to use the spring constant in subsequent parts, use the full precision value you calculated, only rounding as a final step before submitting your answer. Part B After driving a portion of the route, the taptap is fully loaded with a total of 26 people including the driver, with an average mass of 62 kg per persa addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs ha somehow not yet compressed to their maximum amount. How much are the springs compressed? Enter the compression numerically in meters using two significant figures. View Available Hint(s) r = Submit 195) ΑΣΦ Previous Answers X Incorrect; Try Again; 4 attempts remaining Pearson ? m

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**Learning Goal: Understanding Hooke's Law for a Spring**

Hooke's law states that the restoring force \( \vec{F} \) on a spring when it has been stretched or compressed is proportional to the displacement \( \vec{z} \) of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed.

Recall that \( \vec{F} \propto \vec{z} \) means that \( \vec{F} \) is equal to a constant times \( \vec{z} \). For a spring, the proportionality constant is called the spring constant and denoted by \( k \). The spring constant is a property of the spring and must be measured experimentally. The larger the value of \( k \), the stiffer the spring. In equation form, Hooke's law can be written as:

\[ \vec{F} = -k \vec{z} \]

The minus sign indicates that the force is in the opposite direction to that of the spring’s displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by \( F = k z \), where \( z \) is the magnitude of the displacement.

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**Problem Context:**

A 62 kg driver gets into an empty taptap to start the day's work. The springs compress \( 1.8 \times 10^{-2} \) m. What is the effective spring constant of the system in the taptap?

1. **Enter the spring constant numerically in newtons per meter using two significant figures.**
   \[
   k = 3.4 \times 10^4 \, \text{N/m}
   \]

2. **Feedback:**
   - Correct: Ensure to use the spring constant in subsequent parts, using the full precision value calculated, only rounding as a final step before submitting your answer.

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**Part B: Spring Compression with Load**

After driving a portion of the route, the taptap is fully loaded with a total of 26 people including the driver, with an average mass of 62 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum
Transcribed Image Text:**Learning Goal: Understanding Hooke's Law for a Spring** Hooke's law states that the restoring force \( \vec{F} \) on a spring when it has been stretched or compressed is proportional to the displacement \( \vec{z} \) of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed. Recall that \( \vec{F} \propto \vec{z} \) means that \( \vec{F} \) is equal to a constant times \( \vec{z} \). For a spring, the proportionality constant is called the spring constant and denoted by \( k \). The spring constant is a property of the spring and must be measured experimentally. The larger the value of \( k \), the stiffer the spring. In equation form, Hooke's law can be written as: \[ \vec{F} = -k \vec{z} \] The minus sign indicates that the force is in the opposite direction to that of the spring’s displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by \( F = k z \), where \( z \) is the magnitude of the displacement. --- **Problem Context:** A 62 kg driver gets into an empty taptap to start the day's work. The springs compress \( 1.8 \times 10^{-2} \) m. What is the effective spring constant of the system in the taptap? 1. **Enter the spring constant numerically in newtons per meter using two significant figures.** \[ k = 3.4 \times 10^4 \, \text{N/m} \] 2. **Feedback:** - Correct: Ensure to use the spring constant in subsequent parts, using the full precision value calculated, only rounding as a final step before submitting your answer. --- **Part B: Spring Compression with Load** After driving a portion of the route, the taptap is fully loaded with a total of 26 people including the driver, with an average mass of 62 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum
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