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Two objects are connected by a light string that passes over a frictionless pulley and frictionless inclined plane,taking m1=2.0kg,m2=6.0kg and theta =55.0 ,draw a free diagram for m1 and m2,Determine the speed of each object 2.0s after being released from rest

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To analyze the system of two masses connected by a string over a frictionless pulley and a frictionless inclined plane, we will first draw the free body diagrams for both masses and then calculate their acceleration to determine their speed after 2 seconds.

Free Body Diagrams

  1. For m1 (2.0 kg): This mass is on the inclined plane.

    • The gravitational force acting downward is \( F_g = m_1 \cdot g = 2.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 19.62 , \text{N} \).
    • The component of gravitational force acting down the incline is \( F_{\text{down}} = m_1 \cdot g \cdot \sin(\theta) = 2.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot \sin(55^\circ) \).
    • The component of gravitational force acting perpendicular to the incline is \( F_{\text{normal}} = m_1 \cdot g \cdot \cos(\theta) \).
    • The tension \( T \) in the string acts upward along the incline.

  2. For m2 (6.0 kg): This mass hangs vertically.

    • The gravitational force acting downward is \( F_g = m_2 \cdot g = 6.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 58.86 , \text{N} \).
    • The tension \( T \) in the string acts upward.

Equations of Motion

For m1 (on the incline)

Using Newton's second law along the incline: \[ m_1 g \sin(\theta) - T = m_1 a \tag{1} \]

For m2 (hanging vertically)

Using Newton's second law: \[ m_2 g - T = m_2 a \tag{2} \]

Substituting Values

  1. Calculate the gravitational force components:

    • For \( m_1 \):
      • \( F_{\text{down}} = 2.0 \cdot 9.81 \cdot \sin(55^\circ) \approx 2.0 \cdot 9.81 \cdot 0.819 = 16.10 , \text{N} \)
    • For \( m_2 \):
      • \( F_g = 6.0 \cdot 9.81 = 58.86 , \text{N} \)

  2. Substituting \( T \) from (1) into (2): \[ m_2 g - (m_1 g \sin(\theta) - m_1 a) = m_2 a \] \[ 58.86 - 16.10 + 2.0 a = 6.0 a \] \[ 58.86 - 16.10 = 6.0 a - 2.0 a \] \[ 42.76 = 4.0 a \] \[ a = \frac{42.76}{4.0} = 10.69 , \text{m/s}^2 \]

Calculate Speeds After 2 seconds

Starting from rest, using the equation of motion: \[ v = u + at \] Where \( u = 0 \) (initial speed), \( a = 10.69 , \text{m/s}^2 \), and \( t = 2.0 , \text{s} \). \[ v = 0 + (10.69)(2.0) = 21.38 , \text{m/s} \]

Conclusion

  • The acceleration of both objects is \( \approx 10.69 , \text{m/s}^2 \).
  • The speed of each object after 2 seconds is \( \approx 21.38 , \text{m/s} \).

Both objects (m1 moving up the incline and m2 moving downwards) will have the same magnitude of speed.