To find the coefficient of kinetic friction ($\mu_k$) between the crate and the floor, we can use the following physics principles:

1. Since the crate is moving at a constant speed, the net force in the horizontal direction is zero. This means the applied horizontal force is equal to the frictional force: $F_{\text{friction}} = F_{\text{horizontal}}.$

2. The frictional force is given by: $F_{\text{friction}} = \mu_k \cdot F_{\text{normal}}.$

3. The normal force ($F_{\text{normal}}$) can be calculated by summing up the vertical forces. These include:

   • The downward force due to the weight of the crate ($F_{\text{gravity}} = 1320 \, \text{N}$).
   • The vertical component of the applied force ($F_{\text{vertical}} = F \sin \theta$), which adds to the weight because the force is applied at an angle below the horizontal.

Step 1: Break down the forces

The horizontal component of the applied force: $F_{\text{horizontal}} = F \cos \theta = 240 \cos(20^\circ).$

The vertical component of the applied force: $F_{\text{vertical}} = F \sin \theta = 240 \sin(20^\circ).$

The normal force: $F_{\text{normal}} = F_{\text{gravity}} + F_{\text{vertical}} = 1320 + 240 \sin(20^\circ).$

Step 2: Relate forces to find $\mu_k$

The frictional force is equal to the horizontal component of the applied force: $F_{\text{friction}} = F_{\text{horizontal}} = 240 \cos(20^\circ).$

The frictional force is also equal to $\mu_k \cdot F_{\text{normal}}$. Thus: $\mu_k = \frac{F_{\text{friction}}}{F_{\text{normal}}}.$

Substitute the values: $\mu_k = \frac{240 \cos(20^\circ)}{1320 + 240 \sin(20^\circ)}.$

Step 3: Calculate numerical values

• $\cos(20^\circ) \approx 0.9397$,
• $\sin(20^\circ) \approx 0.3420$.

Substitute these: $F_{\text{horizontal}} = 240 \cdot 0.9397 \approx 225.53 \, \text{N}.$ $F_{\text{vertical}} = 240 \cdot 0.3420 \approx 82.08 \, \text{N}.$ $F_{\text{normal}} = 1320 + 82.08 \approx 1402.08 \, \text{N}.$

Now calculate $\mu_k$: $\mu_k = \frac{225.53}{1402.08} \approx 0.161.$

Final Answer:

The coefficient of kinetic friction ($\mu_k$) is approximately: $\boxed{0.161}.$

Let me know if you'd like me to break down any specific step further!

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Relative Questions:

1. How would the coefficient of kinetic friction change if the force was applied at a steeper angle?
2. What role does the constant speed of the crate play in simplifying this problem?
3. How is the normal force affected if the applied force angle is above the horizontal instead of below?
4. How does increasing the weight of the crate affect the coefficient of friction?
5. What happens if the applied force is not sufficient to overcome friction?

Tip: Always resolve forces into their horizontal and vertical components when dealing with inclined or angled forces.