Math Problem Statement

Two blocks, M_1 and M_2, are connected by a massless string that passes over a massless pulley. M_1 has a mass of 2.25 kg and is on an incline of angle θ_1 = 47.5° that has a coefficient of kinetic friction μ_1 = 0.205. M_2 has a mass of 7.35 kg and is on an incline of angle θ_2 = 33.5° that has a coefficient of kinetic friction μ_2 = 0.105. The two-block system is in motion, with block M_2 sliding down the ramp. Find the magnitude a_2 of the acceleration of block M_2.

Solution

This is a classic problem of a system of two masses connected by a pulley, with the masses on inclined planes and experiencing friction. The aim is to calculate the acceleration of block M2M_2. Here's the strategy for solving the problem:

Given:

  • M1=2.25kgM_1 = 2.25 \, \text{kg}
  • M2=7.35kgM_2 = 7.35 \, \text{kg}
  • θ1=47.5\theta_1 = 47.5^\circ, θ2=33.5\theta_2 = 33.5^\circ
  • Coefficient of kinetic friction for M1M_1: μ1=0.205\mu_1 = 0.205
  • Coefficient of kinetic friction for M2M_2: μ2=0.105\mu_2 = 0.105

Steps:

  1. Forces on Block M1M_1:

    • The force due to gravity along the incline: Fg1=M1gsin(θ1)F_{g1} = M_1 g \sin(\theta_1)
    • The friction force on M1M_1: Ffriction1=μ1M1gcos(θ1)F_{\text{friction}_1} = \mu_1 M_1 g \cos(\theta_1)
    • Total net force on M1M_1: FM1=M1gsin(θ1)μ1M1gcos(θ1)F_{M_1} = M_1 g \sin(\theta_1) - \mu_1 M_1 g \cos(\theta_1)
  2. Forces on Block M2M_2:

    • The force due to gravity along the incline: Fg2=M2gsin(θ2)F_{g2} = M_2 g \sin(\theta_2)
    • The friction force on M2M_2: Ffriction2=μ2M2gcos(θ2)F_{\text{friction}_2} = \mu_2 M_2 g \cos(\theta_2)
    • Total net force on M2M_2: FM2=M2gsin(θ2)μ2M2gcos(θ2)F_{M_2} = M_2 g \sin(\theta_2) - \mu_2 M_2 g \cos(\theta_2)
  3. Applying Newton's Second Law: The net forces on both blocks will result in their acceleration, and since they are connected by a pulley, they will have the same magnitude of acceleration aa.

    For block M1M_1: M1a=M1gsin(θ1)μ1M1gcos(θ1)M_1 a = M_1 g \sin(\theta_1) - \mu_1 M_1 g \cos(\theta_1)

    For block M2M_2: M2a=M2gsin(θ2)μ2M2gcos(θ2)M_2 a = M_2 g \sin(\theta_2) - \mu_2 M_2 g \cos(\theta_2)

  4. Equating the forces: Since the tension in the rope is the same for both blocks and they have the same acceleration aa, we solve these two equations simultaneously for aa.

Calculation:

Substitute the given values into the equations:

For block M1M_1: M1a=M1gsin(47.5)μ1M1gcos(47.5)M_1 a = M_1 g \sin(47.5^\circ) - \mu_1 M_1 g \cos(47.5^\circ) 2.25a=2.25×9.8×sin(47.5)0.205×2.25×9.8×cos(47.5)2.25 a = 2.25 \times 9.8 \times \sin(47.5^\circ) - 0.205 \times 2.25 \times 9.8 \times \cos(47.5^\circ)

For block M2M_2: M2a=M2gsin(33.5)μ2M2gcos(33.5)M_2 a = M_2 g \sin(33.5^\circ) - \mu_2 M_2 g \cos(33.5^\circ) 7.35a=7.35×9.8×sin(33.5)0.105×7.35×9.8×cos(33.5)7.35 a = 7.35 \times 9.8 \times \sin(33.5^\circ) - 0.105 \times 7.35 \times 9.8 \times \cos(33.5^\circ)

Let me calculate the values now.The magnitude of the acceleration a2a_2 of block M2M_2 is approximately 2.11m/s22.11 \, \text{m/s}^2.

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

  1. How does the coefficient of kinetic friction affect the motion of the blocks?
  2. What would happen to the acceleration if the incline angles were increased or decreased?
  3. How would adding mass to block M1M_1 affect the system’s acceleration?
  4. What would the acceleration be if the pulley had mass or friction?
  5. How can we calculate the tension in the string?

Tip: Always check if the forces acting parallel and perpendicular to the incline are correctly decomposed into their respective components when solving similar problems.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Dynamics
Inclined Plane
Friction

Formulas

Net force on M_1: M_1 * a = M_1 * g * sin(θ_1) - μ_1 * M_1 * g * cos(θ_1)
Net force on M_2: M_2 * a = M_2 * g * sin(θ_2) - μ_2 * M_2 * g * cos(θ_2)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12