Question

Бомба массой 250 кг свободно падает с высоты 800 м. Чему равна ее потенциальная и кинетическая энергия на высоте 100 м от поверхности Земли.

Answer 1

Дано                 решение Е=mgh₁=250·800·10=2·10⁶Мдж

m-250kг                             Е=mgh₂=250·100·10=250·10³Дж

h-800м

h-100м

g=10

Найти Е                                       Ответ; 1)2Мдж  2)250 Дж

Answer 2

Calculation of Potential and Kinetic Energy

To calculate the potential and kinetic energy of a bomb with a mass of 250 kg freely falling from a height of 800 m, we need to consider the formulas for potential energy and kinetic energy.

The potential energy (PE) of an object at a certain height is given by the formula:

PE = mgh

where: - m is the mass of the object (250 kg in this case) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - h is the height above the reference point (800 m in this case)

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2)mv^2

where: - m is the mass of the object (250 kg in this case) - v is the velocity of the object

To find the velocity of the bomb at a height of 100 m from the surface of the Earth, we can use the principle of conservation of energy. At any given height, the sum of the potential energy and kinetic energy remains constant.

At a height of 800 m, the potential energy is maximum, and the kinetic energy is zero. At a height of 100 m, the potential energy decreases, and the kinetic energy increases.

Let's calculate the potential and kinetic energy at a height of 100 m using the given information.

Calculation:

1. Potential Energy (PE): - PE = mgh - PE = 250 kg * 9.8 m/s^2 * 100 m - PE = 245,000 J

2. Kinetic Energy (KE): - To calculate the kinetic energy, we need to find the velocity of the bomb at a height of 100 m. - Using the principle of conservation of energy, we can equate the potential energy at 800 m to the sum of the potential energy at 100 m and the kinetic energy at 100 m. - PE at 800 m = PE at 100 m + KE at 100 m - 245,000 J = mgh + (1/2)mv^2 - Since the mass and height are constant, we can rearrange the equation to solve for the velocity (v) at 100 m. - (1/2)mv^2 = 245,000 J - mgh - (1/2) * 250 kg * v^2 = 245,000 J - 250 kg * 9.8 m/s^2 * 800 m - v^2 = (245,000 J - 250 kg * 9.8 m/s^2 * 800 m) * 2 / 250 kg - v^2 = 1,960,000 J / 250 kg - v^2 = 7,840 m^2/s^2 - v = √(7,840 m^2/s^2) - v ≈ 88.49 m/s

Now that we have the velocity at 100 m, we can calculate the kinetic energy.

- KE = (1/2)mv^2 - KE = (1/2) * 250 kg * (88.49 m/s)^2 - KE ≈ 976,562.25 J

Answer:

The potential energy of the bomb at a height of 100 m from the surface of the Earth is approximately 245,000 J, and the kinetic energy is approximately 976,562.25 J.

Please note that the calculations are based on the given information and assumptions, and the values are approximate.