Question:

A 2.5 kg box is initially at rest on a horizontal surface. A constant force F of magnitude 10 N is applied horizontally to the box, causing it to accelerate. The coefficient of kinetic friction between the box and the surface is 0.3.

a) Draw a free body diagram of the box showing all the forces acting on it.

b) Calculate the acceleration of the box.

c) Determine the force of kinetic friction acting on the box.

d) If the force F is increased to 15 N, what will be the new acceleration of the box?

Answer:

a) The free body diagram of the box is as follows:

b) To calculate the acceleration of the box, we need to sum the forces acting on it and apply Newton's second law:

The forces acting on the box are the applied force F and the force of kinetic friction. The equation becomes:

Where F is the applied force, f_k is the force of kinetic friction, m is the mass of the box, and a is the acceleration.

c) The force of kinetic friction can be calculated using the equation:

Where $\mu_k$ is the coefficient of kinetic friction and N is the normal force acting on the box. In this case, since the box is on a horizontal surface and is not sinking into the surface, the normal force is equal to the gravitational force acting on the box:

Substituting these values into the equation, we get:

d) To calculate the new acceleration of the box when the force F is increased to 15 N, we can use the equation:

Substituting the new value of F and solving for a:

Now, let's solve the calculations.

Solution:

a) Free body diagram:

b) Applying Newton's second law:

c) Calculating the force of kinetic friction:

d) New acceleration when F is increased to 15 N:

Therefore, the answers to the questions are:

a) b) The acceleration of the box is approximately 1.06 m/s^2. c) The force of kinetic friction acting on the box is approximately 7.35 N. d) The new acceleration of the box, with an increased force F of 15 N, is approximately 3.06 m/s^2.