Introduction
In physics, we are often taught that Newton's Second Law is $F = ma$. However, this is a simplified version that applies to objects with constant mass. When mass is added to or removed from a system, we must return to the more fundamental definition of force: the rate of change of momentum.
The Fundamental Principle
Newton's Second Law, in its most general form, states that force is the time derivative of momentum ($p$):
$$F = \frac{dp}{dt} = \frac{d}{dt}(mv)$$
If the velocity $v$ is kept constant, the equation becomes:
$$F = v \frac{dm}{dt}$$
Where:
- $v$ is the velocity of the system.
- $\frac{dm}{dt}$ is the rate at which mass is added to the system.
Problem Solving: Step-by-Step
Given:
- Initial mass of the vehicle ($M$) = $500\text{ kg}$ (Note: In this specific problem, since the sand falls vertically, its horizontal velocity is zero, and we only care about the rate of mass accumulation).
- Velocity of the vehicle ($v$) = $10\text{ ms}^{-1}$.
- Rate of mass addition ($ rac{dm}{dt}$) = $10\text{ kg/min}$.
Step 1: Convert units to SI units.
- $\frac{dm}{dt} = 10\text{ kg/min} = \frac{10\text{ kg}}{60\text{ s}} = \frac{1}{6}\text{ kg/s}$.
Step 2: Apply the formula. Since we want to keep the velocity constant, the force $F$ applied must counteract the reduction in velocity caused by the sand's inertia (or rather, maintain the momentum of the system).
$$F = v \times \frac{dm}{dt}$$ $$F = 10 \times \frac{1}{6}$$ $$F = \frac{10}{6} = 1.67\text{ N}$$
Conclusion
To keep the vehicle moving at a constant speed of $10\text{ ms}^{-1}$, a force of approximately $1.67\text{ N}$ must be applied. This force is necessary to accelerate the incoming mass of sand from its initial horizontal velocity of $0$ to the vehicle's speed of $10\text{ ms}^{-1}$.