Problem 12 - Physics

Question

A \(30.0\) -kg block is resting on a flat horizontal table. On top of this block is resting a \(15.0\) kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is \(325 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the lower block and the table is \(0.600\), and the coefficient of static friction between the two blocks is \(0.900\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the force \(\overrightarrow{\mathbf{F}}\).

Step-by-Step Solution

Answer

(a) Spring compression is approximately 0.407 m. (b) Force magnitude is approximately 396.9 N.

Step 1: Determine Force of Static Friction

Calculate the maximum static friction force that can act on the upper block before it begins to slip. The force of static friction is given by the formula: \( f_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. Here, \( N = m_2 \cdot g = 15.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \). Calculate \( f_s = 0.900 \times 15.0 \times 9.8 = 132.3 \, \text{N} \).

Step 2: Spring Compression Calculation

When the upper block is on the verge of slipping, the force of static friction equals the restoring force of the spring. The spring force \( f_s = k \cdot x \), gives us the equation: \( 132.3 = 325 \cdot x \). Solve for \( x \) to find the compression of the spring: \( x = \frac{132.3}{325} \approx 0.407 \, \text{m} \).

Step 3: Determine Kinetic Friction Force

Calculate the kinetic friction force that opposes the motion of the lower block on the table using: \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N = (m_1 + m_2) \cdot g = (30.0 + 15.0) \times 9.8 \). Thus, \( N = 45.0 \times 9.8 = 441 \, \text{N} \). Compute \( f_k = 0.600 \times 441 = 264.6 \, \text{N} \).

Step 4: Calculate Force F

The net force acting on the 30 kg block is the applied force \( F \) minus the kinetic friction force \( f_k \). To keep the blocks moving at a constant speed, \( F - f_k = f_s \). Substitute \( f_s \) and \( f_k \) into the equation to find \( F \): \( F - 264.6 = 132.3 \). Solve for \( F \): \( F = 396.9 \, \text{N} \).

Key Concepts

Spring Constant

The spring constant, often denoted as "\(k\)," is a fundamental characteristic of a spring. It quantifies the stiffness of a spring. The higher the spring constant, the stiffer the spring, meaning it requires more force to compress or stretch the spring by a certain amount. Mathematically, the spring constant is represented in the formula for Hooke's Law:

• \(F_s = k \cdot x\)

Where:

• \(F_s\) is the force exerted by the spring, measured in Newtons (N)
• \(k\) is the spring constant, measured in Newtons per meter (N/m)
• \(x\) is the displacement or compression of the spring from its equilibrium position, measured in meters (m)

In the given exercise, the spring constant is 325 N/m, which means it requires 325 Newtons of force to compress the spring by 1 meter. This value is crucial as it directly correlates with how much force is required to achieve a certain compression, as seen in the calculation.Understanding the spring constant helps in various applications, from mechanical devices to measuring how much energy a spring can store.

Kinetic Friction

Kinetic friction is the force resisting the movement of two bodies sliding against each other. It plays a significant role in understanding motion, especially in scenarios involving sliding blocks or cars on the road. Unlike static friction, which acts on objects that are not moving, kinetic friction only applies to objects in motion.The force of kinetic friction \(f_k\) is calculated using the formula:

• \(f_k = \mu_k \cdot N\)

Where:

• \(\mu_k\) is the coefficient of kinetic friction, a dimensionless value representing the ratio between the force of friction and the normal force.
• \(N\) is the normal force, often equivalent to the weight of the object (\(m \cdot g\), where \(m\) is mass and \(g\) is the acceleration due to gravity).

In the problem, the coefficient of kinetic friction \(\mu_k\) is 0.600 for the lower block sliding on the table, and the normal force is calculated based on the masses of both blocks. This force opposes the direction of motion and is essential for determining the net force required to keep the blocks moving at a constant speed.Knowing how to calculate and understand kinetic friction is crucial for solving real-world problems involving motion.

Compression of Spring

The compression of a spring refers to the displacement of the spring from its natural length, typically due to an applied force. This concept is vital in the study of mechanical systems where springs are used for storing potential energy and for absorption of shock.In the given exercise, the compression of the spring \(x\) is related to the maximum static friction force that acts just before slipping occurs.To find the compression, the equation according to Hooke's Law is used:

• \(f_s = k \cdot x\)

Where \(f_s\) is the static friction force found in the problem to be 132.3 N, and \(k\) is the spring constant.Hence, solving for \(x\) gives us:

• \(x = \frac{f_s}{k} = \frac{132.3}{325} \approx 0.407 \, \text{m}\)

This calculation of 0.407 meters tells us how much the spring is compressed from its resting position when the upper block reaches the verge of slipping.Understanding spring compression is critical in designing systems that require resistance to force or changes in their potential energy state.