A body with a constant mass moved uniformly before braking, and at the moment of stopping the braking

| September 11, 2021 | education

A body with a constant mass moved uniformly before braking, and at the moment of stopping the braking force reached the value F = 40 N. Determine the braking force t = 3 s after the start of braking if the braking distance varied with time according to the law S = 196t- t ^ 3

Task data: F (braking force at the moment of stopping) = 40 N; t1 (considered braking time) = 3 s; the law of change in the coordinates of the body: S = 196t – t3.

1) The law of speed change: V = S ‘= (196t – t ^ 3)’ = 196 * 1 – 3 * t ^ 2 = 196 – 3t2.

2) Duration of deceleration: 0 = 196 – 3t ^ 2, whence we express: t = √ (196/3) ≈ 8 s.

3) The law of acceleration change: a = V ‘= (196 – 3t ^ 2)’ = 0 – 3 * 2t = -6t.

4) Acceleration after 8 s: a = -6t = -6 * 8 = -48 m / s2.

5) Body weight: m = F / a = 40/48 = 0.833 kg.

6) Acceleration after 3 s: a1 = -6t1 = -6 * 3 = -18 m / s2.

7) Braking force after 3 s: F1 = m * a1 = 0.833 * 18 ≈ 15 N.



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