College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Transcribed Image Text:**Problem Statement:**
Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the man is 90 kg, and the mass of the woman is 56 kg. The woman pushes on the man with a force of 49 N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.
**Solution Approach:**
1. **Given Data:**
- Mass of the man (\( m_m \)): 90 kg
- Mass of the woman (\( m_w \)): 56 kg
- Force exerted by the woman (\( F \)): 49 N (due east)
2. **Newton’s Second Law:**
- The formula for acceleration: \( a = \frac{F}{m} \)
3. **Determine Acceleration:**
(a) **Acceleration of the Man:**
- Using \( a = \frac{F}{m} \):
- \( a_m = \frac{49 \, \text{N}}{90 \, \text{kg}} \)
- \( a_m = 0.544 \, \text{m/s}^2 \) due east
(b) **Acceleration of the Woman:**
- By Newton’s Third Law, the woman experiences an equal and opposite force.
- \( F \) experienced by the woman is 49 N (due west).
- Using \( a = \frac{F}{m} \):
- \( a_w = \frac{49 \, \text{N}}{56 \, \text{kg}} \)
- \( a_w = 0.875 \, \text{m/s}^2 \) due west
**Conclusion:**
- The acceleration of the man is \( 0.544 \, \text{m/s}^2 \) due east.
- The acceleration of the woman is \( 0.875 \, \text{m/s}^2 \) due west.
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