Question

2. Тело массой 1 кг свободно падает с некоторой высоты. В момент падения на землю его кинетическая энергия равна 100 Дж. С какой высоты падает тело. ​

Answer 1

Ответ: 10м

Объяснение:

Ек=mV²/2

Отсюда

V=√2Ек/m=√2*100/1=10√2м/с

h=V²/2h=200/20=10м

Answer 2

Calculation of the Height of the Falling Object

To calculate the height from which the object falls, we can use the principle of conservation of energy. The initial potential energy of the object is converted into kinetic energy as it falls. At the moment it hits the ground, all of its potential energy is converted into kinetic energy.

Let's denote the mass of the object as m (1 kg in this case), the height from which it falls as h, and the acceleration due to gravity as g (approximately 9.8 m/s^2).

The potential energy of the object at height h is given by mgh, where g is the acceleration due to gravity.

The kinetic energy of the object when it hits the ground is given by (1/2)mv^2, where v is the velocity of the object just before it hits the ground.

According to the principle of conservation of energy, the potential energy at height h is equal to the kinetic energy just before hitting the ground:

mgh = (1/2)mv^2

We are given that the kinetic energy is 100 J, so we can rewrite the equation as:

mgh = 100 J

Substituting the values, we have:

(1 kg)(9.8 m/s^2)(h) = 100 J

Simplifying the equation, we find:

h = 100 J / (1 kg)(9.8 m/s^2)

Calculating the value, we get:

h ≈ 10.2 meters

Therefore, the object falls from a height of approximately 10.2 meters.