This problem involves a cart experiencing an applied force and friction, which eventually slows it down. The goal is to find the acceleration, net force, frictional force, and the coefficient of friction.

Let's break it down step by step:

Given:

• Mass of the cart $m = 16 \, \text{kg}$
• Applied force $F_{\text{applied}} = 50 \, \text{N}$
• Initial velocity $v_0 = 9 \, \text{m/s}$
• Final velocity $v = 0 \, \text{m/s}$ (since the cart stops)
• Distance traveled before stopping $d = 8 \, \text{m}$

Step 1: Find the acceleration

Using the kinematic equation: $v^2 = v_0^2 + 2a d$ Substitute the known values: $0 = (9)^2 + 2a(8)$ $0 = 81 + 16a$ $16a = -81$ $a = \frac{-81}{16} = -5.06 \, \text{m/s}^2$

So, the acceleration is $a = -5.06 \, \text{m/s}^2$.

Step 2: Calculate the net force

The net force $F_{\text{net}}$ acting on the cart can be found using Newton’s second law: $F_{\text{net}} = ma$ $F_{\text{net}} = 16 \, \text{kg} \times (-5.06 \, \text{m/s}^2)$ $F_{\text{net}} = -80.96 \, \text{N}$ So, the net force is $F_{\text{net}} = -80.96 \, \text{N}$ (negative because it acts in the opposite direction of motion).

Step 3: Determine the frictional force

The frictional force $F_{\text{friction}}$ is the force that causes the cart to stop. The net force is the combination of the applied force and friction. Therefore: $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$ Solving for $F_{\text{friction}}$: $F_{\text{friction}} = F_{\text{applied}} - F_{\text{net}}$ $F_{\text{friction}} = 50 \, \text{N} - (-80.96 \, \text{N})$ $F_{\text{friction}} = 50 \, \text{N} + 80.96 \, \text{N}$ $F_{\text{friction}} = 130.96 \, \text{N}$

So, the frictional force is $F_{\text{friction}} = 130.96 \, \text{N}$.

Step 4: Find the coefficient of friction

The frictional force is related to the normal force by: $F_{\text{friction}} = \mu F_{\text{normal}}$ For a flat surface, $F_{\text{normal}} = mg$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity. Therefore: $F_{\text{friction}} = \mu mg$ Solving for $\mu$: $\mu = \frac{F_{\text{friction}}}{mg}$ $\mu = \frac{130.96 \, \text{N}}{16 \, \text{kg} \times 9.8 \, \text{m/s}^2}$ $\mu = \frac{130.96}{156.8}$ $\mu = 0.835$

So, the coefficient of friction is $\mu = 0.835$.

Summary of Results:

• Acceleration: $a = -5.06 \, \text{m/s}^2$
• Net force: $F_{\text{net}} = -80.96 \, \text{N}$
• Frictional force: $F_{\text{friction}} = 130.96 \, \text{N}$
• Coefficient of friction: $\mu = 0.835$

Would you like more details or further explanation on any part of this problem?

Here are some related questions you could explore next:

1. How does changing the mass of the cart affect the acceleration in this scenario?
2. What would happen to the frictional force if the applied force were doubled?
3. How can we calculate the work done by friction to stop the cart?
4. What is the effect of increasing the coefficient of friction on the stopping distance?
5. How does the angle of inclination of the surface influence the net force acting on the cart?

Tip: Always ensure that units are consistent throughout the calculations, particularly when using kinematic equations or applying Newton’s second law!