Let's go through the question step by step. Here's what the problem asks:

Given:

• A student pushes a 4 kg box initially at rest along a rough, flat table with a rightward force of 10 N.
• The coefficient of sliding friction between the box and the table is $\mu = 0.2$.

You need to:

1. Draw and label all forces acting on the box on a Free Body Diagram.
2. Calculate the force of friction.
3. Calculate the acceleration of the box.

a. Free Body Diagram

There are four forces acting on the box:

1. Applied force ($F_{applied}$): A rightward force of 10 N.
2. Gravitational force ($F_g$): The weight of the box, $F_g = m \cdot g$, where $m = 4 \text{ kg}$ and $g = 9.8 \text{ m/s}^2$.
3. Normal force ($F_N$): This force is perpendicular to the surface and balances the gravitational force.
4. Frictional force ($F_{friction}$): Acts leftward, opposing the motion.

Here is the calculation for some of these forces:

• $F_g = m \cdot g = 4 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N}$.
• $F_N = F_g$, because the box is on a flat surface with no vertical motion, so the normal force exactly balances the gravitational force.

The free-body diagram should include:

• $10 \, \text{N}$ force to the right (applied force).
• $F_{friction}$ to the left.
• $39.2 \, \text{N}$ downward (gravitational force).
• $39.2 \, \text{N}$ upward (normal force).

b. Calculate Force of Friction

The force of friction $F_{friction}$ is given by the equation: $F_{friction} = \mu \cdot F_N$ Where:

• $\mu = 0.2$ (coefficient of friction),
• $F_N = 39.2 \, \text{N}$.

$F_{friction} = 0.2 \cdot 39.2 = 7.84 \, \text{N}$

c. Calculate Acceleration

To calculate the acceleration, use Newton's second law: $F_{net} = m \cdot a$ Where:

• $F_{net}$ is the net force acting on the box,
• $m = 4 \, \text{kg}$,
• $a$ is the acceleration.

The net force $F_{net}$ is the applied force minus the frictional force: $F_{net} = 10 \, \text{N} - 7.84 \, \text{N} = 2.16 \, \text{N}$

Now, using Newton's second law: $a = \frac{F_{net}}{m} = \frac{2.16 \, \text{N}}{4 \, \text{kg}} = 0.54 \, \text{m/s}^2$

Summary:

• Frictional force: $7.84 \, \text{N}$
• Acceleration: $0.54 \, \text{m/s}^2$

Would you like a further explanation or details about any part?

Here are 5 related questions you might be interested in:

1. How would the result change if the surface were smoother (lower coefficient of friction)?
2. What happens to the acceleration if the mass of the box is doubled?
3. How do you calculate frictional force for an inclined plane?
4. Can you explain the difference between static and kinetic friction?
5. What would happen if there were no friction at all?

Tip: Always verify the direction of forces when drawing Free Body Diagrams, as it helps avoid mistakes when calculating net forces.