A 5.1- kg concrete block rests on a level table. A 3.4- kg mass is attached to the block by a string passing over a light, frictionless pulley. If the acceleration of the block is measured to be 1.3 m/s2, what is the coefficient of friction between the block and the table? What is the acceleration of the wooden block? What is the tension in the string during the acceleration? Determine the acceleration for the above situation when the coefficient of friction between the block and the table is 0.27. What is the tension when the coefficient of friction between the block and the table is 0.27?
axdas
To find the coefficient of friction between the block and the table, we can start by considering the forces acting on the system.
1. Identify the forces:
- Weight of the block (acting downwards): Fₓ = m₁ * g
- Tension in the string (acting upwards): Fᵧ
- Force of friction (opposing motion): Fᶠ
- Weight of the hanging mass (acting downwards): F₂ = m₂ * g
2. Apply Newton's second law (F = m * a) to the block:
- Sum of forces in the horizontal direction: Fₓ - Fᶠ = m₁ * a
- Sum of forces in the vertical direction: Fᵧ - F₁ = m₁ * g
- The only unknown in the horizontal equation is the force of friction Fᶠ.
3. Determine the acceleration of the wooden block:
- We are given that the acceleration of the block is 1.3 m/s². Therefore, a = 1.3 m/s².
4. Substitute values into the equations:
- Fₓ - Fᶠ = m₁ * a
- Fᵧ - F₁ = m₁ * g
Plugging in the given values:
- m₁ = 5.1 kg (mass of the block)
- m₂ = 3.4 kg (mass hanging from the block)
- a = 1.3 m/s² (acceleration)
- g = 9.8 m/s² (acceleration due to gravity)
- Fₓ - Fᶠ = m₁ * a -> (1)
- Fᵧ - F₁ = m₁ * g -> (2)
5. Solve for Fᶠ using equation (1):
- Fₓ - Fᶠ = m₁ * a
- 5.1 kg * 1.3 m/s² - Fᶠ = 5.1 kg * 1.3 m/s²
- Fᶠ = 5.1 kg * 1.3 m/s² - 5.1 kg * 1.3 m/s²
- Fᶠ = 0 N
From this, we can see that the force of friction is zero. Hence, the coefficient of friction between the block and the table is also zero.
6. Determine the tension in the string during acceleration:
- Fᵧ - F₁ = m₁ * g
- Fᵧ - (5.1 kg * 9.8 m/s²) = 5.1 kg * 9.8 m/s²
- Fᵧ - 50.04 N = 50.04 N
- Fᵧ = 50.04 N + 50.04 N
- Fᵧ = 100.08 N
Therefore, the tension in the string during acceleration is 100.08 N.
7. Determine the acceleration when the coefficient of friction is 0.27:
- The force of friction can now be determined using the equation Fᶠ = μ * N, where μ is the coefficient of friction and N is the normal force between the block and the table.
- The normal force N is equal to the weight of the block, N = m₁ * g.
- Fᶠ = μ * m₁ * g
Plugging in the given values:
- μ = 0.27 (coefficient of friction)
- m₁ = 5.1 kg (mass of the block)
- g = 9.8 m/s² (acceleration due to gravity)
- Fᶠ = 0.27 * 5.1 kg * 9.8 m/s²
8. Determine the tension when the coefficient of friction is 0.27:
- Fᵧ - F₁ - Fᶠ = m₁ * g
- Fᵧ - (5.1 kg * 9.8 m/s²) - (0.27 * 5.1 kg * 9.8 m/s²) = 5.1 kg * 9.8 m/s²
- Fᵧ - 50.04 N - (0.27 * 50.04 N) = 50.04 N
- Fᵧ = 50.04 N + 0.27 * 50.04 N
- Fᵧ = 50.04 N + 13.5112 N
- Fᵧ ≈ 63.5512 N
Therefore, the tension in the string when the coefficient of friction is 0.27 is approximately 63.5512 N.