It seems like you're dealing with a problem involving motion and forces. From the details you've given:

• Mass ($m$) = 1200 kg
• Coefficient of friction ($\mu$) = 0.20
• Initial velocity ($v_i$) = 30 m/s
• Final velocity ($v_f$) = 0 m/s (since it stops)

The problem likely requires you to calculate the distance before the object stops due to friction.

Approach:

The stopping distance can be found using the work-energy principle. The work done by the friction force is equal to the change in kinetic energy of the object.

1. Frictional Force:
   The frictional force $f_{\text{friction}}$ is given by: $f_{\text{friction}} = \mu \cdot m \cdot g$ where $g$ is the acceleration due to gravity ($g \approx 9.8 \, \text{m/s}^2$).

2. Kinetic Energy:
   The initial kinetic energy ($KE_{\text{initial}}$) of the object is: $KE_{\text{initial}} = \frac{1}{2} m v_i^2$

3. Work Done by Friction:
   The work done by friction is: $W = f_{\text{friction}} \cdot d$ where $d$ is the stopping distance.

4. Energy Conservation:
   The work done by friction equals the change in kinetic energy. Since the final velocity is zero (the object stops), we have: $KE_{\text{initial}} = W$ $\frac{1}{2} m v_i^2 = \mu \cdot m \cdot g \cdot d$

5. Solving for Distance:
   We can now solve for the distance $d$: $d = \frac{v_i^2}{2 \mu g}$

Plugging in the values:

• $v_i = 30 \, \text{m/s}$
• $\mu = 0.20$
• $g = 9.8 \, \text{m/s}^2$

$d = \frac{30^2}{2 \times 0.20 \times 9.8}$

Let me calculate that for you.

Calculation:

$d = \frac{900}{2 \times 0.20 \times 9.8}$ $d = \frac{900}{3.92}$ $d \approx 229.59 \, \text{meters}$

Answer:

The object will stop after approximately 229.6 meters.

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Would you like a more detailed explanation of any steps or formulas? Feel free to ask!

Here are some additional questions related to your topic:

1. How does the coefficient of friction affect the stopping distance?
2. What role does the object's mass play in determining the stopping distance?
3. How would the stopping distance change if the initial velocity were doubled?
4. What other forces, besides friction, could affect the stopping distance?
5. How does Universal Gravitation relate to everyday motion, like stopping an object?

Tip: When solving physics problems involving friction, always check whether the problem involves kinetic or static friction, as their values can differ!