Question

Высота наклонной плоскости равна 1,2 м , а длина 10,8 м . Для подъёма на этой наклонной плоскости груза массой 180 кг потребовалась сила 250 Н . Определите КПД наклонной плоскости и силу трения. помогите найти силу трения

Answer 1

1) 10,8 м / 1,2 м = 9 выигрыш в силе, которую даёт нам наклонная плоскость.Р=F= 250 H   КПД= mgh / F = 180 кг* 10 Н/кг * 10,8 м / 250 = 77,76

Answer 2

Calculating Efficiency of the Inclined Plane and Frictional Force

To calculate the efficiency of the inclined plane and the frictional force, we can use the following information:

- Height of the inclined plane (h) = 1.2 m - Length of the inclined plane (l) = 10.8 m - Force required to lift the load (F) = 250 N - Mass of the load (m) = 180 kg

Calculating Efficiency (η)

The efficiency (η) of the inclined plane can be calculated using the formula:

η = (Output Work/Input Work) × 100%

The input work is the work done to lift the load vertically, and the output work is the work done to lift the load along the inclined plane.

The input work can be calculated using the formula: Input Work = Force × Distance = mgh

Where: - m = mass of the load - g = acceleration due to gravity (approximately 9.81 m/s^2) - h = height of the inclined plane

The output work can be calculated using the formula: Output Work = Force × Distance

Let's calculate the input work and output work using the given values and then find the efficiency.

Calculating Frictional Force

The frictional force (Ff) can be calculated using the formula: Ff = μ × N

Where: - μ = coefficient of friction - N = normal force

The normal force (N) can be calculated using the formula: N = m × g

Let's calculate the frictional force using the given values.

Solution

Calculating Input Work: Input Work = Force × Distance = mgh Input Work = 180 kg × 9.81 m/s^2 × 1.2 m Input Work = 2116.16 J

Calculating Output Work: Output Work = Force × Distance Output Work = 250 N × 10.8 m Output Work = 2700 J

Calculating Efficiency (η): η = (Output Work/Input Work) × 100% η = (2700 J / 2116.16 J) × 100% η ≈ 127.6%

Calculating Frictional Force: N = m × g N = 180 kg × 9.81 m/s^2 N ≈ 1765.8 N

Ff = μ × N 250 N = μ × 1765.8 N μ ≈ 0.1414

Therefore, the efficiency of the inclined plane is approximately 127.6% and the frictional force is approximately 0.1414.

I hope this helps! If you have further questions, feel free to ask.