Understanding Power in Mechanics
Power is defined as the rate at which work is done or the rate at which energy is transferred. For an object moving at a constant velocity $v$, the power $P$ required to maintain that motion against a resisting force $F$ is given by:
$$P = F \cdot v$$
In this problem, the train must overcome two forces to maintain a constant speed up an incline:
- Frictional force ($F_f$): Given as $1.28 \times 10^4 \text{ N}$.
- Gravitational component ($F_g$): The force component acting down the incline due to gravity, calculated as $mg \sin(\theta)$.
Step-by-Step Solution
1. Identify Given Data
- Mass ($m$) = $2 \times 10^5 \text{ kg}$
- Speed ($v$) = $72 \text{ km/h} = 72 \times \frac{1000}{3600} = 20 \text{ m/s}$
- Frictional force ($F_f$) = $1.28 \times 10^4 \text{ N}$
- Incline slope: $1 \text{ m}$ rise per $100 \text{ m}$ along the slope, so $\sin(\theta) = \frac{1}{100} = 0.01$
- Gravity ($g$) $\approx 9.8 \text{ m/s}^2$
2. Calculate Total Opposing Force ($F_{total}$)
To move at a constant speed, the engine must exert a force equal to the sum of the opposing forces: $$F_{total} = F_f + mg \sin(\theta)$$ $$F_{total} = 1.28 \times 10^4 + (2 \times 10^5)(9.8)(0.01)$$ $$F_{total} = 12,800 + 19,600 = 32,400 \text{ N}$$
3. Calculate Power
$$P = F_{total} \cdot v$$ $$P = 32,400 \text{ N} \times 20 \text{ m/s}$$ $$P = 648,000 \text{ W} = 6.48 \times 10^5 \text{ W} = 648 \text{ kW}$$
Summary
The train engine must develop 648 kW of power to maintain its speed against friction and the incline.