| Surface | Frictional Coefficient |
|---|---|
| Gravel and dirt road | 0.35 |
| Wet grassy field | 0.20 |
| Dry asphalt | 0.60 |
| Wet asphalt | 0.45 |
| Snow-covered road | 0.20 – 0.30 |
| Ice | 0.05 – 0.10 |
| Dry concrete | 0.70 |
| Wet concrete | 0.60 |
Example:
What was the initial velocity of a car that skidded first 50 feet on dry concrete and then 90 feet over a grassy pasture, given the following equation and friction coefficients?
Equation: v = √[2g (f1s1 + f2s2)]
Friction coefficients:
- f1 (dry concrete) = 0.7
- f2 (grassy pasture) = 0.2
Distances:
- s1 (dry concrete) = 50 feet
- s2 (grassy pasture) = 90 feet
Note: g = 32.2 ft/sec² (acceleration due to gravity)
v = √[2(32.2 ft/sec²) (0.7 × 50 ft + 0.2 × 90 ft)]
v = √[2(32.2 ft/sec²) (35 ft + 18 ft)]
v = √[2(32.2 ft/sec²) (53 ft)]
v = √[2(32.2 ft/sec²)(53 ft)]
v = √[3421.2 ft²/sec²]
v = √[3421.2 ft²/sec²] v ≈ 58.5 ft/sec
Convert miles into mph
1 mile = 5280 feet; 1 hour = 3600 seconds
v ≈ 58.5 ft/sec * (1 mile / 5280 feet) * (3600 seconds / 1 hour) v ≈ 39.9 mph
The initial velocity of the car was approximately 40 mph.