Question

Тело массой m = 0,2 кг соскальзывает с высоты H = 8 м по наклонной плоскости, плавно переходящей в вертикальную петлю радиусом R = 2 м. Определить работу силы трения при движении тела до верхней точки петли, если давление тела на петлю в верхней точке N равно 2 Н.

Answer 1

Дано:                                        Решение
h=8м                                    Aтр=mgH-2,5mgR-0,5RN
m=0,2кг                                  Aтр=H -2,5-0,5×N=8-2,5-0,5×2=4,5Дж
N=2H
R=2м

Answer 2

Problem Analysis

We are given the following information: - Mass of the body, m = 0.2 kg - Height, H = 8 m - Radius of the loop, R = 2 m - Pressure on the loop at the top, N = 2 N

We need to determine the work done by the force of friction as the body moves to the top of the loop.

Solution

To solve this problem, we can break it down into two parts: 1. The body sliding down the inclined plane. 2. The body moving in the vertical loop.

Let's calculate the work done in each part separately.

Part 1: Sliding down the inclined plane

When the body is sliding down the inclined plane, the work done by the force of gravity is given by the formula:

Work = Force × Distance × cos(θ)

In this case, the force is the component of the gravitational force parallel to the inclined plane, which is given by:

Force = m × g × sin(θ)

The distance is the height of the inclined plane, H.

The angle, θ, between the inclined plane and the horizontal is given by:

θ = tan^(-1)(H/R)

Let's calculate the work done in this part.

Using the given values: - Mass, m = 0.2 kg - Height, H = 8 m - Radius, R = 2 m

We can calculate the angle θ:

θ = tan^(-1)(8/2) = tan^(-1)(4) ≈ 75.96°

The force is given by:

Force = m × g × sin(θ)

Using the approximate value of the acceleration due to gravity, g = 9.8 m/s^2, we can calculate the force:

Force = 0.2 × 9.8 × sin(75.96°) ≈ 18.94 N

The work done is given by:

Work = Force × Distance × cos(θ)

Substituting the values:

Work = 18.94 × 8 × cos(75.96°) ≈ 18.94 × 8 × 0.258 ≈ 39.07 J

Therefore, the work done in the first part is approximately 39.07 J.

Part 2: Moving in the vertical loop

In this part, the force of friction does not do any work because it is perpendicular to the displacement of the body. Therefore, the work done by the force of friction is zero.

Total work done

The total work done is the sum of the work done in both parts.

Total work done = Work in Part 1 + Work in Part 2 = 39.07 J + 0 J = 39.07 J

Therefore, the work done by the force of friction as the body moves to the top of the loop is approximately 39.07 J.

Answer

The work done by the force of friction as the body moves to the top of the loop is approximately 39.07 J.