Question

Тело массой 2 кг падает с высоты 20 м из состояния покоя и в момент падения на землю имеет скорость 15 м\с. Определите среднюю силу сопротивления воздуха при падении тела.Помогите, пожалуйста, очень нужно!

Answer 1

$2Sa=V^2-V_{0}^2$

$2S(g-a)=V^2$

$F=ma$

$a=\frac{F}{m}$

$g-\frac{F}{m}=\frac{V^2}{2S}$

$gm-\frac{mV^2}{2S}=F$

$F\approx 20-11,25$

$F\approx 8,75H$

Answer 2

Problem Analysis

We are given the following information: - Mass of the body (m) = 2 kg - Initial height (h0) = 20 m - Initial velocity (v0) = 0 m/s - Final velocity (v) = 15 m/s

We need to determine the average air resistance force (F) acting on the body during its fall.

Solution

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the body is conserved during its fall, neglecting any other forces except air resistance.

The total mechanical energy (E) of the body is given by the sum of its potential energy (PE) and kinetic energy (KE):

E = PE + KE

At the initial position, the body has only potential energy, given by:

PE = mgh0

At the final position, the body has only kinetic energy, given by:

KE = (1/2)mv^2

Since the total mechanical energy is conserved, we have:

PE + KE = Constant

Substituting the expressions for PE and KE, we get:

mgh0 + (1/2)mv^2 = Constant

Simplifying the equation, we have:

gh0 + (1/2)v^2 = Constant

Since the body is falling freely, the constant term is zero. Therefore, we have:

gh0 + (1/2)v^2 = 0

Solving for g, we get:

g = -(1/2)(v^2)/h0

The force of air resistance (F) acting on the body is given by:

F = mg - ma

where m is the mass of the body, g is the acceleration due to gravity, and a is the acceleration of the body.

Since the body is falling freely, the acceleration due to gravity is equal to the acceleration of the body. Therefore, we have:

F = m(g - a)

To find the average air resistance force, we need to determine the average acceleration (a) of the body during its fall. We can use the kinematic equation:

v^2 = v0^2 + 2aΔs

where v is the final velocity, v0 is the initial velocity, a is the acceleration, and Δs is the displacement.

Since the body starts from rest (v0 = 0), the equation simplifies to:

v^2 = 2aΔs

Solving for a, we get:

a = v^2 / (2Δs)

Substituting the given values, we have:

a = (15^2) / (2 * 20)

Now we can calculate the average air resistance force (F) using the formula:

F = m(g - a)

Substituting the values of m, g, and a, we get:

F = 2 * (-(1/2)(15^2)/20 - (15^2) / (2 * 20))

Calculating the value of F, we find:

F = -22.5 N

Therefore, the average air resistance force acting on the body during its fall is -22.5 N.

Answer

The average air resistance force acting on the body during its fall is -22.5 N.