Question:

A 10 kg box is placed on a rough horizontal surface with a coefficient of friction of 0.4. A force of 50 N is applied to the box at an angle of 30 degrees from the horizontal. Draw a free-body diagram of the box showing all the forces acting on it and find the net force and the acceleration of the box.

Answer:

Free-body diagram of the box:

        Normal force (N)
           /|
          / |
         /  |
       /    |
      /      |  Weight (mg)
     F        |
   /          |
 /            |
/_____________|
 Frictional force (f)

The forces acting on the box are the weight (mg), the normal force (N), the force applied (F), and the frictional force (f).

To find the net force, we need to resolve the applied force into its components. The component of the force in the horizontal direction is Fcos(30°) = 50N * cos(30°) = 43.3N. The component in the vertical direction is Fsin(30°) = 50N * sin(30°) = 25N.

Calculating the frictional force, f = μ*N = 0.4 * N. Given that N = mg, we can substitute N in terms of m and g.

f = 0.4 * mg f = 0.4 * 10kg * 9.8m/s^2 f = 39.2N

Now, we can calculate the net force:

Net force in the x-direction = Fcosθ - f Net force in the x-direction = 43.3N - 39.2N = 4.1N

Net force in the y-direction = N - W Net force in the y-direction = N - mg Net force in the y-direction = 0N

Now, we can find the net force using Pythagoras theorem:

Net force = √((Net force in the x-direction)^2 + (Net force in the y-direction)^2) Net force = √((4.1N)^2 + (0N)^2) Net force = 4.1N

The net force acts in the horizontal direction. Since the net force is non-zero, the box will accelerate in the direction of the net force. We can calculate the acceleration using the equation:

Acceleration = Net force / mass Acceleration = 4.1N / 10kg Acceleration = 0.41 m/s^2

So, the net force acting on the box is 4.1N and the acceleration of the box is 0.41 m/s^2.