Math Problem Statement

A spring with a force constant of 110 N/mN/m is used to push a 0.30-kgkg block of wood against a wall, as shown in the figure (Figure 1).Find the minimum compression of the spring needed to keep the block from falling, given that the coefficient of static friction between the block and the wall is 0.42.

Solution

To solve this problem, we need to analyze the forces acting on the block and the spring in equilibrium.

Step 1: Forces Acting on the Block

The block is pushed against the wall by the spring, and the goal is to prevent the block from falling. Here's a breakdown of the forces involved:

  • Weight (Gravitational Force): Fg=mgF_g = m \cdot g

    • Where mm is the mass of the block (0.30 kg), and gg is the acceleration due to gravity (9.8 m/s²).
  • Normal Force (N): The spring applies a force that acts perpendicular to the wall. This force is the normal force NN, which is equal to the restoring force of the spring: N=Fspring=kxN = F_{\text{spring}} = k \cdot x

    • Where k=110N/mk = 110 \, \text{N/m} is the spring constant, and xx is the compression of the spring.
  • Frictional Force: The static friction prevents the block from sliding down. The maximum frictional force is: Ffriction=μsNF_{\text{friction}} = \mu_s \cdot N

    • Where μs=0.42\mu_s = 0.42 is the coefficient of static friction.

Step 2: Equilibrium Condition

To keep the block from falling, the frictional force must be at least equal to the gravitational force acting downward: FfrictionFgF_{\text{friction}} \geq F_g This gives us the condition: μsNmg\mu_s \cdot N \geq m \cdot g

Substituting N=kxN = k \cdot x, we have: μskxmg\mu_s \cdot k \cdot x \geq m \cdot g

Step 3: Solve for xx (Minimum Compression)

Now, solving for xx: xmgμskx \geq \frac{m \cdot g}{\mu_s \cdot k}

Substitute the known values: x(0.30kg)(9.8m/s2)(0.42)(110N/m)x \geq \frac{(0.30 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2)}{(0.42) \cdot (110 \, \text{N/m})}

Step 4: Calculation

Let me compute that:The minimum compression of the spring needed to keep the block from falling is approximately 0.064m0.064 \, \text{m} or 6.4 cm.

Would you like further details or have any other questions?

Here are 5 related questions you might find interesting:

  1. What happens if the coefficient of static friction is increased or decreased?
  2. How would the spring compression change if the mass of the block increased?
  3. How would this scenario change if the spring constant were different?
  4. What is the relationship between the angle of the surface and the required compression?
  5. Can the same analysis apply if the block is on an inclined plane?

Tip: Static friction always adjusts to resist motion up to its maximum value, determined by the normal force and the coefficient of static friction.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Static Friction
Hooke's Law
Force Equilibrium

Formulas

F_g = m * g
N = k * x
F_friction = μ_s * N
μ_s * k * x ≥ m * g
x ≥ (m * g) / (μ_s * k)

Theorems

Newton's First Law of Motion (Equilibrium)
Hooke's Law

Suitable Grade Level

High School Physics