Problem 85 - Physics

Question

Problem A \(0.80-\mathrm{kg}\) box is attached to a horizontal spring whose spring constant is \(59 \mathrm{~N}\) \(\mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving?

Step-by-Step Solution

Answer

The spring can be stretched 0.0983 meters without the box moving.

Step 1: Understanding the Problem

We need to determine how far the spring can be stretched while keeping the box stationary. The box will not move as long as the static frictional force is greater than or equal to the spring's restoring force.

Step 2: Calculate Maximum Static Friction Force

The maximum static friction force can be calculated using the formula \[ F_{s} = \mu_{s} \times N \]where \( N \) is the normal force. For this problem, the normal force equals the weight of the box, calculated as \[ N = m \times g = 0.80 \times 9.8 = 7.84 \text{ N} \]Thus, the maximum static friction force is \[ F_{s} = 0.74 \times 7.84 = 5.8016 \text{ N} \]

Step 3: Calculate Force from Spring

The force exerted by the spring when stretched by \( x \) meters is given by Hooke's Law: \[ F = k \times x \]where \( k = 59 \text{ N/m} \). Our goal is to find \( x \) when the spring force equals the maximum static friction force.

Step 4: Equate Forces and Solve for Displacement

Set the static friction force equal to the spring force to prevent movement: \[ k \times x = F_{s} \]Substituting the known values gives us \[ 59 \times x = 5.8016 \]Solving for \( x \) we get \[ x = \frac{5.8016}{59} = 0.0983 \text{ meters} \]

Step 5: Conclusion

The spring can be stretched up to approximately 0.0983 meters without the box moving.

Key Concepts

Spring Constant

The spring constant (\( k \)) is a measure of a spring's stiffness. It is given in units of Newtons per meter (N/m). This constant determines how much force is needed to stretch or compress the spring by one unit of length. Given in the problem, the spring constant is 59 N/m. This means for every meter you stretch or compress the spring, you need to exert 59 Newtons of force. Factors such as the material of the spring and its dimensions can affect its spring constant. A higher spring constant indicates a stiffer spring. In practical scenarios, understanding the spring constant helps predict how a spring behaves under different loads. This is crucial in applications ranging from vehicle suspensions to the buttons on a calculator.

Hooke's Law

Hooke's Law is a principle that states that the force exerted by a spring is proportional to the displacement or deformation experienced by the spring from its rest position. This can be mathematically expressed as: \[ F = k \times x \] Where:

• \( F \) is the force exerted by the spring (in Newtons)
• \( k \) is the spring constant (in N/m)
• \( x \) is the displacement of the spring from its natural length (in meters)

Hooke's Law is applicable as long as the deformation doesn't surpass the elastic limit of the material. Beyond this point, permanent deformation may occur, and Hooke's Law no longer applies. In our exercise, we use Hooke's Law to determine the force exerted by the spring, which is essential for comparing it to the static friction force.

Normal Force

The normal force is a contact force exerted by a surface perpendicularly to an object resting on it. It is an essential element in problems involving friction. By balancing forces, the normal force can be determined, often equalling the weight of the object when resting on a horizontal surface. In our exercise, the box resting on the table experiences a normal force equal to its weight. This force can be calculated as the product of mass (\( m \)) and gravitational acceleration (\( g \)), or \( N = m \times g \). The gravitational acceleration is approximately 9.8 m/s². The box has a mass of 0.80 kg, leading to a normal force of \( 7.84 \text{ N} \). This normal force is vital for determining the maximum static friction force.

Maximum Static Friction Force

The maximum static friction force is the greatest amount of force that a surface can exert to prevent an object from sliding. It is calculated using the static friction coefficient \( \mu_s \) and the normal force \( N \): \[ F_s = \mu_s \times N \] In our problem, the static friction coefficient \( \mu_s \) is 0.74, and the normal force is 7.84 N. So, the maximum static friction force is \( 5.8016 \text{ N} \). This maximum force defines the threshold beyond which the box will start moving if a greater force is applied. In our scenario, as long as the spring's restoring force (due to its deformation) does not exceed this static friction force, the box stays stationary. Understanding this concept is crucial in calculating the maximum stretch allowed for the spring without causing movement.