Math Problem Statement

Two blocks are connected by a string, as shown in the figure (Figure 1). The smooth inclined surface makes an angle of 35∘ with the horizontal, and the block on the incline has a mass of 5.7 kg. The mass of the hanging block is m = 4.1 kg. What is the acceleration of the hanging block?

Solution

To determine the acceleration of the hanging block, we need to apply Newton's second law of motion to both blocks and solve for the unknown acceleration. Here's how we approach the problem step by step:

Step 1: Set up the forces

  • For the block on the incline (mass = 5.7 kg), there are two forces to consider:

    1. Gravitational force acting vertically downward, Fg=m1g=5.7kg×9.8m/s2=55.86NF_g = m_1 g = 5.7 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 55.86 \, \text{N}.
    2. The force of gravity along the incline: F=m1gsin(θ)=5.7×9.8×sin(35)F_{\parallel} = m_1 g \sin(\theta) = 5.7 \times 9.8 \times \sin(35^\circ).
  • For the hanging block (mass = 4.1 kg), there are:

    1. Gravitational force acting vertically downward, Fg,2=m2g=4.1×9.8=40.18NF_{g,2} = m_2 g = 4.1 \times 9.8 = 40.18 \, \text{N}.

Step 2: Net force and tension in the string

Let TT be the tension in the string, and let aa be the acceleration of both blocks (they move together, so they have the same acceleration).

  • For the block on the incline: m1a=Tm1gsin(θ)m_1 a = T - m_1 g \sin(\theta)

  • For the hanging block: m2a=m2gTm_2 a = m_2 g - T

Step 3: Combine the equations

Now we have two equations:

  1. m1a=Tm1gsin(θ)m_1 a = T - m_1 g \sin(\theta)
  2. m2a=m2gTm_2 a = m_2 g - T

We can solve this system by adding the two equations to eliminate TT: m1a+m2a=m2gm1gsin(θ)m_1 a + m_2 a = m_2 g - m_1 g \sin(\theta) Factor out aa on the left side: a(m1+m2)=m2gm1gsin(θ)a (m_1 + m_2) = m_2 g - m_1 g \sin(\theta) Now, solve for aa: a=m2gm1gsin(θ)m1+m2a = \frac{m_2 g - m_1 g \sin(\theta)}{m_1 + m_2}

Step 4: Plug in the values

Substitute the known values:

  • m1=5.7kgm_1 = 5.7 \, \text{kg}
  • m2=4.1kgm_2 = 4.1 \, \text{kg}
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2
  • θ=35\theta = 35^\circ

a=(4.1×9.8)(5.7×9.8×sin(35))5.7+4.1a = \frac{(4.1 \times 9.8) - (5.7 \times 9.8 \times \sin(35^\circ))}{5.7 + 4.1} a=40.185.7×9.8×0.57369.8a = \frac{40.18 - 5.7 \times 9.8 \times 0.5736}{9.8} a=40.1832.059.8a = \frac{40.18 - 32.05}{9.8} a=8.139.8a = \frac{8.13}{9.8} a0.83m/s2a \approx 0.83 \, \text{m/s}^2

Final Answer:

The acceleration of the hanging block is approximately 0.83m/s20.83 \, \text{m/s}^2.


Would you like further details on any of the steps or concepts? Here are some related questions for further practice:

  1. How would the acceleration change if the angle of inclination were increased?
  2. How would the tension in the string compare for the two blocks?
  3. What would happen if the surface were not frictionless?
  4. How do you solve for acceleration in a system with more than two blocks?
  5. How can you find the speed of the blocks after a certain time?

Tip: To solve problems like this, always break the forces into components along the direction of motion (usually parallel and perpendicular to the incline).

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law of Motion
Free Body Diagrams
Forces on Inclined Planes

Formulas

F = ma
F_parallel = m * g * sin(θ)
a = (m2 * g - m1 * g * sin(θ)) / (m1 + m2)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-11