Question

Вагон массой 20 т движется равнозамедленно с ускорением 0,3 м/с 2 и начальной скоростью 54 км/ч. Найти силу торможения, действующую на вагон, время движения вагона до остановки и перемещение вагона.

Answer 1

Ответ:

Fт = 6 кН;    t₁ = 50 c;     s₁ = 375 м;

Объяснение:

m = 20 т = 20 000 кг

a = 0.3 м/с²

v₀ = 54 км/ч = 15 м/с

Fт - ? - сила торможения

t₁ - ? - время торможения до остановки

s₁ - ? - тормозной путь

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Fт = ma = 20 000 · 0.3 = 6 000 (H) = 6 КН

t₁ = v₀/a = 15 : 0.3 = 50 (c)

s₁ = v₀t₁ - 0.5at₁² = 15 · 50 - 0.5 · 0.3 · 2500 = 375 (м)

Answer 2

Problem Analysis

We are given the following information: - Mass of the wagon (m) = 20 t = 20000 kg - Acceleration (a) = 0.3 m/s^2 - Initial velocity (v0) = 54 km/h

We need to find: - Braking force acting on the wagon - Time taken for the wagon to stop - Displacement of the wagon during this time

Calculation

To find the braking force (F), we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: F = m * a Substituting the given values: F = 20000 kg * 0.3 m/s^2 = 6000 N

The braking force acting on the wagon is 6000 N.

To find the time taken for the wagon to stop, we can use the following kinematic equation: v = v0 + at where: - v is the final velocity (which is 0 m/s since the wagon stops) - v0 is the initial velocity - a is the acceleration - t is the time taken

Rearranging the equation to solve for time (t): t = (v - v0) / a

Substituting the given values: t = (0 m/s - 54 km/h) / (0.3 m/s^2)

Converting the initial velocity from km/h to m/s: 54 km/h = 54 * (1000 m / 3600 s) = 15 m/s

Substituting the values: t = (0 m/s - 15 m/s) / 0.3 m/s^2 = -15 m/s / 0.3 m/s^2 = -50 s

The negative sign indicates that the wagon is decelerating. However, time cannot be negative in this context, so we take the absolute value: t = 50 s

The time taken for the wagon to stop is 50 seconds.

To find the displacement of the wagon during this time, we can use the following kinematic equation: s = v0 * t + (1/2) * a * t^2 where: - s is the displacement - v0 is the initial velocity - a is the acceleration - t is the time taken

Substituting the given values: s = 15 m/s * 50 s + (1/2) * 0.3 m/s^2 * (50 s)^2

Simplifying the equation: s = 750 m + 0.15 m/s^2 * 2500 s^2 = 750 m + 375 m = 1125 m

The displacement of the wagon during this time is 1125 meters.

Summary

Based on the given information, we have determined the following: - The braking force acting on the wagon is 6000 N. - The time taken for the wagon to stop is 50 seconds. - The displacement of the wagon during this time is 1125 meters.

Please let me know if there is anything else I can help you with!