Question

Вантаж масою 100 кг тягнуть рівномірно вгору з силою 600 Н вздовж похилої площини з кутом нахилу 30° до горизонту. З яким прискоренням цей вантаж буде рухатися вниз, якщо його відпустити?

Answer 1

When a 100 kg object is pulled upwards with a force of 600 N along an inclined plane with an angle of 30° to the horizontal, we can calculate the acceleration of the object when it is released.

To find the acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

In this case, the force pulling the object upwards is 600 N, and the mass of the object is 100 kg. The force of gravity acting on the object can be split into two components: one parallel to the inclined plane and one perpendicular to it. The component of the force of gravity parallel to the inclined plane is given by mg sinθ, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of inclination.

The net force acting on the object is the difference between the force pulling it upwards and the force of gravity acting downwards. So we have:

Net force = Force pulling upwards - Force of gravity = 600 N - (mg sinθ)

Now we can use Newton's second law to find the acceleration:

ma = 600 N - (mg sinθ)

Simplifying the equation, we have:

a = (600 N - mg sinθ) / m

Substituting the values, we get:

a = (600 N - (100 kg)(9.8 m/s^2)(sin 30°)) / 100 kg

Evaluating the expression, we find:

a ≈ 2.55 m/s^2

Therefore, when the 100 kg object is released, it will accelerate downwards with an approximate acceleration of 2.55 m/s^2.

[[1]]