Question

Помогите пожалуйста. Небольшое тело массой 200 кг равномерно поднимают по наклонной плоскости длиной 40 метров, образующей угол 30º с горизонтом, прикладывая силу 1,5кН вдоль линии движения. Сколько времени будет соскальзывать это тело с вершины наклонной плоскости из состояния покоя, если его отпустить?

Answer 1

Ответ:

Решение.

Рис.1. Движение тела вверх Рис.2. Движение тела вниз Рассмотрим движение тела вверх (рис. 1). Тело движется равномерно, то есть его ускорение равно 0. Запишем для тела второй закон Ньютона в векторной форме и в проекциях на оси и : ; Сила трения пропорциональна нормальному давлению:

Answer 2

To calculate the time it takes for the body to slide down the inclined plane, we can use the principles of Newtonian mechanics. The key concept we need to apply is the conservation of mechanical energy.

Given information:

- Mass of the body (m) = 200 kg - Length of the inclined plane (s) = 40 meters - Angle of the inclined plane (θ) = 30 degrees - Applied force (F) = 1.5 kN = 1500 N

Approach:

1. Calculate the gravitational force acting on the body. 2. Determine the component of the gravitational force parallel to the inclined plane. 3. Calculate the net force acting on the body. 4. Use the work-energy principle to find the change in kinetic energy. 5. Equate the change in kinetic energy to the work done by the net force. 6. Solve for the time it takes for the body to slide down.

Calculation:

1. The gravitational force acting on the body is given by the formula: F_gravity = m * g where g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the given values, we have: F_gravity = 200 kg * 9.8 m/s^2 = 1960 N [[1]].

2. The component of the gravitational force parallel to the inclined plane is given by: F_parallel = F_gravity * sin(θ) Substituting the given values, we have: F_parallel = 1960 N * sin(30 degrees) = 980 N [[2]].

3. The net force acting on the body is the sum of the applied force and the component of the gravitational force parallel to the inclined plane: F_net = F_applied + F_parallel Substituting the given values, we have: F_net = 1500 N + 980 N = 2480 N.

4. According to the work-energy principle, the work done by the net force is equal to the change in kinetic energy of the body. The work done is given by: Work = F_net * s * cos(θ) where s is the length of the inclined plane and θ is the angle of the inclined plane. Substituting the given values, we have: Work = 2480 N * 40 m * cos(30 degrees) = 99200 J [[3]].

5. The change in kinetic energy is given by: ΔKE = KE_final - KE_initial Since the body starts from rest, the initial kinetic energy (KE_initial) is zero. Therefore, the change in kinetic energy is equal to the final kinetic energy (KE_final). Equating the work done to the change in kinetic energy, we have: Work = ΔKE = KE_final Substituting the calculated value of work, we have: 99200 J = KE_final.

6. The final kinetic energy can be calculated using the formula: KE_final = (1/2) * m * v^2 where v is the final velocity of the body. Rearranging the equation, we have: v^2 = (2 * KE_final) / m Substituting the known values, we have: v^2 = (2 * 99200 J) / 200 kg = 992 m^2/s^2.

To find the final velocity, we take the square root of both sides: v = √(992 m^2/s^2) ≈ 31.49 m/s.

Finally, we can calculate the time it takes for the body to slide down using the formula: t = s / v Substituting the known values, we have: t = 40 m / 31.49 m/s ≈ 1.27 seconds.

Therefore, it will take approximately 1.27 seconds for the body to slide down from the top of the inclined plane when released from rest.

Please note that this calculation assumes no friction or air resistance acting on the body.