Question

контейнер масою 50 кг стоїть на горизонтальній площині. Чи рухатись контейнер якщо до нього прикласти горизонтальну силу 2кН? коефіцієнт тертя дорівнює 0,3​

Answer 1

Ответ:

Тело будет двигаться

Объяснение:

F=2кН=2000Н

u=0,3

m=50 кг

g=10м/с2

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Найдём максимальную силу трение Fт:

Fт=uN

По второму закону ньютона N=mg

Fт=umg=0,3*50*10=150н

F>Fт, следовательно тело будет двигаться

Answer 2

Situation Analysis

A 50 kg container is placed on a horizontal surface, and a horizontal force of 2 kN is applied to it. The coefficient of friction is 0.3.

Solution

The first step is to calculate the maximum static friction force that can act on the container. This can be done using the formula:

Maximum Static Friction Force (F_max) = coefficient of friction (μ) × Normal force (N)

Where: - coefficient of friction (μ) = 0.3 - Normal force (N) = mass (m) × acceleration due to gravity (g)

Calculation

The normal force (N) can be calculated as: N = m × g Where: - m = 50 kg (mass of the container) - g = 9.81 m/s^2 (acceleration due to gravity)

Substitute the values: N = 50 kg × 9.81 m/s^2

Now, calculate the maximum static friction force (F_max): F_max = 0.3 × (50 kg × 9.81 m/s^2)

Result

After calculating, we find that the maximum static friction force (F_max) is equal to 147.15 N.

Conclusion

Since the applied force of 2 kN (2000 N) is greater than the maximum static friction force of 147.15 N, the container will start to move when the force is applied.

Therefore, the container will move when a horizontal force of 2 kN is applied to it.