Please help. Thanks.?
2016-08-02 9:03 pm
In what distance can a 1500kg automobile be stopped if the brake is applied when the speed is 20m/s and the coefficient of sliding friction is 0.7 between the tyre and the ground
回答 (2)
2016-08-02 9:12 pm
Take g = 9.81 m/s²
Initial K.E. = (1/2)mv² = (1/2) × 1500 × 20² = 300000 J
Frictional force, f = μmg = 0.7 × 1500 × 9.81 = 10300 N
Work done by frictional force = f s = 10300s J
Work done by frictional force = Decrease in K.E.
10300s = 300000
Distance = 300000/10300 = 29 m
Initial K.E. = (1/2)mv² = (1/2) × 1500 × 20² = 300000 J
Frictional force, f = μmg = 0.7 × 1500 × 9.81 = 10300 N
Work done by frictional force = f s = 10300s J
Work done by frictional force = Decrease in K.E.
10300s = 300000
Distance = 300000/10300 = 29 m
2016-08-02 10:27 pm
The friction force causes the car’s velocity to decrease from 20 m/s to 0 m/s. Use the following equation to determine the distance.
vf^2 = vi^2 + 2 * a * d
a = -μ * g = -0.7 * 9.8 = -6.86 m/s^2
0 = 20^2 + 2 * -6.86 * d
13.72 * d = 400
d = 400 ÷ 13.72
The distance is approximately 29 meters. I hope this is helpful for you.
vf^2 = vi^2 + 2 * a * d
a = -μ * g = -0.7 * 9.8 = -6.86 m/s^2
0 = 20^2 + 2 * -6.86 * d
13.72 * d = 400
d = 400 ÷ 13.72
The distance is approximately 29 meters. I hope this is helpful for you.
收錄日期: 2021-04-18 15:22:42
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