若一架质量为m=30000kg的航载机以V0=288km/h的速度降落在航母甲板上,飞机滑行s=160m便能停下,求 ①
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(1)

设 V0 = 288km/h = 80 m/s,停下后 V1 = 0 m/s,质量 m = 30000 kg,S = 160 m,

由 V0² - V1² = 2aS,可得 6400 - 0 =2 × a × 160,解得 a = 20 m/s².

(2)

S = 0.5 × a × t ²,代入数据可得 160 = 0.5 × 20 × t ²,

得出 t = 4 s

(3)

根据 F = m × a,可得 F = 30000 × 20 = 6 000 00 N.