17. I A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?

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**Physics Problems for Educational Exercise**

**Problem 17:**
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force?

**Problem 18:**
A fast pitch softball player does a “windmill” pitch, illustrated...

(Note: The image cuts off the rest of Problem 18's description, so further details are not available.)

---

**Detailed Explanation for Problem 17:**

To solve this problem, we need to determine the net force acting on the car as it moves in a circular path.

1. **Given Data:**
   - Mass of the car, \( m = 1500 \) kg
   - Diameter of the circular track, \( d = 200 \) m
   - Speed of the car, \( v = 25 \) m/s

2. **Determine the Radius:**
   - Radius, \( r = \frac{d}{2} = \frac{200}{2} = 100 \) m

3. **Centripetal Force Calculation:**
   The net force on the car moving in a circular path is the centripetal force, which can be calculated using the formula:
   \[
   F_c = \frac{mv^2}{r}
   \]
   Substituting the given values:
   \[
   F_c = \frac{1500 \times (25)^2}{100} = \frac{1500 \times 625}{100} = 9375 \text{ N}
   \]

4. **Direction of the Force:**
   The direction of the net force is towards the center of the circular path, which is the direction of the centripetal force.

5. **Cause of the Force:**
   This force is caused by the frictional force between the car's tires and the track, which provides the necessary centripetal force to keep the car moving in a circular path.

---

**Note:** This section contains three problems from a set, each designed to enhance the understanding of dynamics and forces in circular motion for students.
Transcribed Image Text:**Physics Problems for Educational Exercise** **Problem 17:** A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car? What causes this force? **Problem 18:** A fast pitch softball player does a “windmill” pitch, illustrated... (Note: The image cuts off the rest of Problem 18's description, so further details are not available.) --- **Detailed Explanation for Problem 17:** To solve this problem, we need to determine the net force acting on the car as it moves in a circular path. 1. **Given Data:** - Mass of the car, \( m = 1500 \) kg - Diameter of the circular track, \( d = 200 \) m - Speed of the car, \( v = 25 \) m/s 2. **Determine the Radius:** - Radius, \( r = \frac{d}{2} = \frac{200}{2} = 100 \) m 3. **Centripetal Force Calculation:** The net force on the car moving in a circular path is the centripetal force, which can be calculated using the formula: \[ F_c = \frac{mv^2}{r} \] Substituting the given values: \[ F_c = \frac{1500 \times (25)^2}{100} = \frac{1500 \times 625}{100} = 9375 \text{ N} \] 4. **Direction of the Force:** The direction of the net force is towards the center of the circular path, which is the direction of the centripetal force. 5. **Cause of the Force:** This force is caused by the frictional force between the car's tires and the track, which provides the necessary centripetal force to keep the car moving in a circular path. --- **Note:** This section contains three problems from a set, each designed to enhance the understanding of dynamics and forces in circular motion for students.
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