Trường hợp a: Bỏ qua ma sát

Lực kéo của động cơ

Theo định luật II Newton:

F_k = m a = 2000 \times 0.4 = 800 \text{ N}

Công của động cơ

A = F_k \cdot s

Với quãng đường s tính theo công thức chuyển động thẳng biến đổi đều:

s = v_0 t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 0.4 \times 15^2 = 45 \text{ m}

A = 800 \times 45 = 36.000 \text{ J}

Công suất trung bình

P = \frac{A}{t} = \frac{36.000}{15} = 2400 \text{ W} = 2.4 \text{ kW}

Trường hợp b: Có ma sát

Lực ma sát:

F_{\text{ms}} = \mu mg = 0.05 \times 2000 \times 10 = 1000 \text{ N}

Lực kéo của động cơ

F_k = F_{\text{ms}} + ma = 1000 + 800 = 1800 \text{ N}

Công của động cơ

A = F_k \cdot s = 1800 \times 45 = 81.000 \text{ J}

Công suất trung bình

P = \frac{A}{t} = \frac{81.000}{15} = 5400 \text{ W} = 5.4 \text{ kW}

Kết luận:

Trường hợp a (bỏ qua ma sát):

• Lực kéo: 800 N

• Công: 36.000 J

• Công suất: 2.4 kW

Trường hợp b (có ma sát \mu = 0.05)

• Lực kéo: 1800 N

• Công: 81.000 J

• Công suất: 5.4 kW

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