Question

Для подъема по наклонной плоскости груза массой 1,2 кг была приложена сила 5,4 Н. Высота наклонной плоскости 0,2 м, длина 0,8 м. Определите КПД этой наклонной плоскости. даю 55 балов​

Answer 1

Ответ: ≈0,56=56 %.

Объяснение:

При подъёме груза была произведена работа A=F*s, где F=5,4 Н и s=0,8 м. Отсюда A=5,4*0,8=4,32 Дж. При этом груз приобрёл потенциальную энергию E=m*g*h. где m=1,2 кг, g=10 м/с², h=0,2 м. Отсюда E=1,2*10*0,2=2,4 Дж. Тогда КПД плоскости η=E/A=2,4/4,32≈0,56.

Answer 2

Calculation of Efficiency (КПД) of the Inclined Plane

To calculate the efficiency (КПД) of the inclined plane, we need to determine the work input and work output.

Given information: - Mass of the load (m): 1.2 kg - Applied force (F): 5.4 N - Height of the inclined plane (h): 0.2 m - Length of the inclined plane (l): 0.8 m

Work input (W_in): The work input is the product of the applied force and the distance over which the force is applied. In this case, the distance is the length of the inclined plane.

W_in = F * l

Work output (W_out): The work output is the product of the weight of the load and the height of the inclined plane.

W_out = m * g * h

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Efficiency (КПД): Efficiency is defined as the ratio of the work output to the work input, multiplied by 100% to express it as a percentage.

КПД = (W_out / W_in) * 100%

Now let's calculate the efficiency.

Calculation:

W_in = 5.4 N * 0.8 m = 4.32 JW_out = 1.2 kg * 9.8 m/s^2 * 0.2 m = 2.352 J КПД = (2.352 J / 4.32 J) * 100% ≈ 54.44%

Therefore, the efficiency (КПД) of this inclined plane is approximately 54.44%.

Please note that the calculation assumes ideal conditions and does not take into account factors such as friction or other losses that may affect the actual efficiency.