Questions: A block of mass m₁=40 kg on a horizontal surface is connected to a mass m₂=17.3 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. The coefficient of kinetic friction between m₁ and the horizontal surface is 0.24. (a) What is the magnitude of the acceleration (in m / s²) of the hanging mass? □ m / s² (b) Determine the magnitude of the tension (in N) in the cord above the hanging mass. □ N

A block of mass m₁=40 kg on a horizontal surface is connected to a mass m₂=17.3 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. The coefficient of kinetic friction between m₁ and the horizontal surface is 0.24. (a) What is the magnitude of the acceleration (in m / s²) of the hanging mass? □ m / s² (b) Determine the magnitude of the tension (in N) in the cord above the hanging mass. □ N
Transcript text: A block of mass $m_{1}=40 \mathrm{~kg}$ on a horizontal surface is connected to a mass $m_{2}=17.3 \mathrm{~kg}$ that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. The coefficient of kinetic friction between $m_{1}$ and the horizontal surface is 0.24 . (a) What is the magnitude of the acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) of the hanging mass? $\square$ $\mathrm{m} / \mathrm{s}^{2}$ (b) Determine the magnitude of the tension (in N ) in the cord above the hanging mass. $\square$ N
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Solution

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Solution Steps

Step 1: Identify the forces acting on each mass

  • For \( m_1 \) (40 kg) on the horizontal surface:

    • Tension \( T \) in the string (to the right)
    • Kinetic friction \( f_k \) (to the left)
    • Normal force \( N \) (upward)
    • Gravitational force \( m_1 g \) (downward)

  • For \( m_2 \) (17.3 kg) hanging vertically:

    • Tension \( T \) in the string (upward)
    • Gravitational force \( m_2 g \) (downward)
Step 2: Write the equations of motion

  • For \( m_1 \): \[ T - f_k = m_1 a \] where \( f_k = \mu_k N \) and \( N = m_1 g \), so \( f_k = \mu_k m_1 g \).

  • For \( m_2 \): \[ m_2 g - T = m_2 a \]

Step 3: Solve for acceleration \( a \)

  • Substitute \( f_k \) into the equation for \( m_1 \): \[ T - \mu_k m_1 g = m_1 a \]

  • Combine the equations for \( m_1 \) and \( m_2 \): \[ m_2 g - T = m_2 a \] \[ T = m_2 g - m_2 a \]

  • Substitute \( T \) from the second equation into the first equation: \[ m_2 g - m_2 a - \mu_k m_1 g = m_1 a \] \[ m_2 g - \mu_k m_1 g = m_1 a + m_2 a \] \[ g (m_2 - \mu_k m_1) = a (m_1 + m_2) \] \[ a = \frac{g (m_2 - \mu_k m_1)}{m_1 + m_2} \]

  • Plug in the given values (\( g = 9.8 \, \text{m/s}^2 \), \( m_1 = 40 \, \text{kg} \), \( m_2 = 17.3 \, \text{kg} \), \( \mu_k = 0.24 \)): \[ a = \frac{9.8 \times (17.3 - 0.24 \times 40)}{40 + 17.3} \] \[ a = \frac{9.8 \times (17.3 - 9.6)}{57.3} \] \[ a = \frac{9.8 \times 7.7}{57.3} \] \[ a \approx 1.32 \, \text{m/s}^2 \]

Step 4: Determine the tension \( T \)
  • Use the equation for \( m_2 \): \[ T = m_2 g - m_2 a \] \[ T = 17.3 \times 9.8 - 17.3 \times 1.32 \] \[ T = 169.54 - 22.836 \] \[ T \approx 146.70 \, \text{N} \]

Final Answer

  • (a) The magnitude of the acceleration of the hanging mass is approximately \( 1.32 \, \text{m/s}^2 \).
  • (b) The magnitude of the tension in the cord above the hanging mass is approximately \( 146.70 \, \text{N} \).
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