Math Problem Statement

A block of mass 6 kg resting on a horizontal surface is connected by a cord passing over a light, frictionless pulley to a hanging block of mass 4 kg. The coefficient of kinetic friction between the block and the horizontal surface is 0.5. After the blocks are released find: (a) the acceleration of each block (b) The tension on the cord

Solution

To solve this problem, let’s denote the following variables for convenience:

  • m1=6kgm_1 = 6 \, \text{kg}: mass of the block on the horizontal surface.
  • m2=4kgm_2 = 4 \, \text{kg}: mass of the hanging block.
  • μk=0.5\mu_k = 0.5: coefficient of kinetic friction between m1m_1 and the horizontal surface.
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2: acceleration due to gravity.

Step 1: Identify the Forces on Each Block

For Block m1m_1 on the Horizontal Surface:

  • Gravitational force: Fg1=m1g=69.8=58.8NF_{g1} = m_1 \cdot g = 6 \cdot 9.8 = 58.8 \, \text{N}.
  • Normal force (equal to gravitational force since it's on a horizontal surface): N=58.8NN = 58.8 \, \text{N}.
  • Frictional force opposing motion: Ff=μkN=0.558.8=29.4NF_f = \mu_k \cdot N = 0.5 \cdot 58.8 = 29.4 \, \text{N}.
  • Tension in the cord: TT (this is unknown and is the same on both blocks since the pulley is light and frictionless).

For Block m2m_2 (the Hanging Block):

  • Gravitational force: Fg2=m2g=49.8=39.2NF_{g2} = m_2 \cdot g = 4 \cdot 9.8 = 39.2 \, \text{N}.
  • Tension in the cord: TT.

Step 2: Set Up the Equations of Motion

The system accelerates in the direction of the hanging block, so we can apply Newton’s Second Law for each block.

For m1m_1:

TFf=m1aT - F_f = m_1 \cdot a Substituting Ff=29.4NF_f = 29.4 \, \text{N}: T29.4=6a(Equation 1)T - 29.4 = 6a \quad \text{(Equation 1)}

For m2m_2:

m2gT=m2am_2 \cdot g - T = m_2 \cdot a Substituting Fg2=39.2NF_{g2} = 39.2 \, \text{N}: 39.2T=4a(Equation 2)39.2 - T = 4a \quad \text{(Equation 2)}

Step 3: Solve for Acceleration and Tension

Add Equations 1 and 2 to eliminate TT: (T29.4)+(39.2T)=6a+4a(T - 29.4) + (39.2 - T) = 6a + 4a 9.8=10a9.8 = 10a a=9.810=0.98m/s2a = \frac{9.8}{10} = 0.98 \, \text{m/s}^2

Now, substitute a=0.98m/s2a = 0.98 \, \text{m/s}^2 into Equation 1 to find TT: T29.4=6×0.98T - 29.4 = 6 \times 0.98 T29.4=5.88T - 29.4 = 5.88 T=5.88+29.4=35.28NT = 5.88 + 29.4 = 35.28 \, \text{N}

Final Answers:

  • (a) The acceleration of each block is 0.98m/s20.98 \, \text{m/s}^2.
  • (b) The tension on the cord is 35.28N35.28 \, \text{N}.

Would you like more details, or do you have any questions about this problem?

Related Questions

  1. How would changing the coefficient of kinetic friction affect the acceleration?
  2. What would happen if the mass of m2m_2 were increased?
  3. How does the tension in the cord compare to the weight of m1m_1 and m2m_2?
  4. If the pulley were not frictionless, how would it affect the system?
  5. What would the acceleration be if the surface were frictionless?

Tip:

In pulley problems, always start by analyzing forces on each mass separately and set up equations for each block using Newton's Second Law.

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Frictional Force
Pulley Systems
Equilibrium of Forces

Formulas

F = ma
Frictional force F_f = μ_k * N
Gravitational force F_g = m * g
Tension and acceleration equations for pulley systems

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12