Consider the diagram below of a box with a mass of 6.00 kg. The applied force acts an an angle =20.0o and has a magnitude of FA=85.0 N. A frictional force of 21.0 N acts opposing the motion. What is the acceleration of the box? Give your answer in m/s2 to 2 decimal places. Do NOT include units in your answer.

1 answer

To find the acceleration of the box, we can start by calculating the net force acting on it. The net force \( F_{net} \) can be found by considering the applied force, the angle of the applied force, and the frictional force acting on the box.

  1. Calculate the horizontal component of the applied force \( F_A \): \[ F_{A_x} = F_A \cdot \cos(\theta) \] where:

    • \( F_A = 85.0 , N \)
    • \( \theta = 20.0^\circ \)

    Using \( \cos(20^\circ) \) (which is approximately 0.9397): \[ F_{A_x} = 85.0 , N \cdot \cos(20^\circ) \approx 85.0 , N \cdot 0.9397 \approx 79.91 , N \]

  2. Determine the net force. The net force \( F_{net} \) acting on the box is given by: \[ F_{net} = F_{A_x} - F_{friction} \] where:

    • \( F_{friction} = 21.0 , N \)

    So, \[ F_{net} = 79.91 , N - 21.0 , N \approx 58.91 , N \]

  3. Calculate the acceleration using Newton's second law \( F = ma \): \[ a = \frac{F_{net}}{m} \] where \( m = 6.00 , kg \): \[ a = \frac{58.91 , N}{6.00 , kg} \approx 9.82 , m/s^2 \]

Thus, the acceleration of the box is approximately 9.82.