Math Problem Statement

answer to 4.4 and 4.5

Solution

To solve questions 4.4 and 4.5 from the image, we need to consider the forces acting on both blocks, as well as the acceleration of the system. Here's the breakdown of the problem:

Given Data:

  • Mass of Block 1 (m1m_1): 4 kg
  • Mass of Block 2 (m2m_2): 8 kg
  • Inclination Angle (θ\theta): 30°
  • Coefficient of Kinetic Friction (μk\mu_k): 0.2

Question 4.4: Calculate the magnitude of the acceleration of the system.

Step-by-Step Solution:

  1. Identify Forces on Each Block:

    • For block 1 (m1=4m_1 = 4 kg) on the incline:

      • Weight component along the incline: m1gsinθm_1g \sin \theta
      • Friction force: f1=μkN1f_1 = \mu_k N_1, where N1=m1gcosθN_1 = m_1g \cos \theta (normal force)
    • For block 2 (m2=8m_2 = 8 kg) hanging vertically:

      • Gravitational force: m2gm_2g
  2. Equations for Block 1 (on the incline):

    • Weight component along the incline: m1gsinθ=4×9.8×sin(30°)=19.6Nm_1g \sin \theta = 4 \times 9.8 \times \sin(30°) = 19.6 \, \text{N}
    • Normal force: N1=m1gcosθ=4×9.8×cos(30°)=33.94NN_1 = m_1g \cos \theta = 4 \times 9.8 \times \cos(30°) = 33.94 \, \text{N}
    • Friction force: f1=μkN1=0.2×33.94=6.79Nf_1 = \mu_k N_1 = 0.2 \times 33.94 = 6.79 \, \text{N}
  3. Net Force on Block 1: F1,net=m1gsinθf1=19.66.79=12.81NF_{1,\text{net}} = m_1g \sin \theta - f_1 = 19.6 - 6.79 = 12.81 \, \text{N}

  4. Net Force on Block 2:

    • The force due to gravity: F2,net=m2g=8×9.8=78.4NF_{2,\text{net}} = m_2g = 8 \times 9.8 = 78.4 \, \text{N}
  5. Total Net Force on the System: Fnet=F2,netF1,net=78.412.81=65.59NF_{\text{net}} = F_{2,\text{net}} - F_{1,\text{net}} = 78.4 - 12.81 = 65.59 \, \text{N}

  6. Total Mass of the System: mtotal=m1+m2=4+8=12kgm_{\text{total}} = m_1 + m_2 = 4 + 8 = 12 \, \text{kg}

  7. Acceleration of the System: a=Fnetmtotal=65.5912=5.47m/s2a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{65.59}{12} = 5.47 \, \text{m/s}^2

Answer to 4.4: The magnitude of the acceleration of the system is 5.47 m/s2^2.

Question 4.5: How will the acceleration compare if the positions of the 44 kg block and 88 kg block are switched?

If we switch the positions of the blocks, the larger mass (88 kg) will be on the incline, and the smaller mass (44 kg) will be hanging. The friction force on the 88 kg block will be higher due to its larger normal force, which will decrease the acceleration of the system because the net force driving the motion will be smaller.

Answer to 4.5: The acceleration of the system will be LESS THAN when the 44 kg block is on the incline and the 88 kg block is hanging.


Would you like to know more details or have any questions?

Additional Questions:

  1. What is Newton's Second Law of Motion, and how is it applied in this scenario?
  2. How would the acceleration change if the coefficient of friction was different?
  3. What are the effects of the angle of inclination on the acceleration of the system?
  4. How can we calculate the tension in the string connecting the two blocks?
  5. How would the system's behavior change if there were no friction?

Tip: Understanding the role of friction is crucial in mechanics. It always acts opposite to the direction of motion and reduces the net force, thereby affecting acceleration.

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Friction
Force Analysis

Formulas

Newton's Second Law: F = ma

Theorems

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Suitable Grade Level

High School