**Problem Statement:**It is weigh-in time for the local under 85 kg rugby team. The bathroom scale that is used to assess eligibility can be described by Hooke's law, which is depressed 0.73 cm for its maximum load of 123 kg. What is the scale's effective spring constant \( k \)?\[ k = \underline{\hspace{3cm}} \, \text{N/m} \]A player stands on the scale, and it depresses by 0.49 cm. What is the mass \( m \) of the player?\[ m = \underline{\hspace{3cm}} \, \text{kg} \]**Question Credit:** OpenStax College Physics---**Explanation:**- **Hooke's Law:** This principle is used to describe the behavior of springs. It states that the force needed to compress or extend a spring by some distance is proportional to that distance. Mathematically, it's expressed as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring.- **Objective:** 1. Determine the spring constant \( k \) when the scale is depressed by 0.73 cm under a load of 123 kg. 2. Calculate the mass \( m \) of a player when the scale depresses by 0.49 cm.

Name: **Problem Statement:**It is weigh-in time for the local under 85 kg r
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Description: **Problem Statement:**It is weigh-in time for the local under 85 kg rugby team. The bathroom scale that is used to assess eligibility can be described by Hooke's law, which is depressed 0.73 cm for its maximum load of 123 kg. What is the scale's eff

Given Data: The mass of a maximum load is given as, m=123 kg The depression in the scale is given as, x=0.73 cm The expression for the spring force is given as, F=kx ∴From Newton's second law, F=mgmg=kx.................(1) Here, m is the mass, g is the gravitational acceleration, k is the spring constant, and x is the depression. To calculate the scale's effective spring constant, substitute the values in equation (1) 123 kg×9.81 m/s2=k×0.73 cm×10-2 m1 cm1206.63 kg·m/s2=k×0.73 cm×10-2 m1 cmk=1206.63 kg·m/s20.73 cm×10-2 m1 cmk=165.29×103 N/mNow, if the scale depressed by 0.49 cm, then the mass of the player will be, m×9.81 m/s2=165.29×103 N/m×0.49 cm×10-2 m1 cmm=165.29×103 N/m×0.49 cm×10-2 m1 cm9.81 m/s2m=82.56 kg Therefore, the scale's effective spring constant is 165.29×103 N/m and the mass of a player is 82.56 kg.

Answered: k = N/m A player stands on the scale,…

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k = N/m A player stands on the scale, and it depresses by 0.49 cm. What is the mass m of the player? m = kg

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**Problem Statement:**

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale that is used to assess eligibility can be described by Hooke's law, which is depressed 0.73 cm for its maximum load of 123 kg. What is the scale's effective spring constant \( k \)?

\[ k = \underline{\hspace{3cm}} \, \text{N/m} \]

A player stands on the scale, and it depresses by 0.49 cm. What is the mass \( m \) of the player?

\[ m = \underline{\hspace{3cm}} \, \text{kg} \]

**Question Credit:** OpenStax College Physics

---

**Explanation:**

- **Hooke's Law:** This principle is used to describe the behavior of springs. It states that the force needed to compress or extend a spring by some distance is proportional to that distance. Mathematically, it's expressed as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring.

- **Objective:**
1. Determine the spring constant \( k \) when the scale is depressed by 0.73 cm under a load of 123 kg.
2. Calculate the mass \( m \) of a player when the scale depresses by 0.49 cm.

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