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Car with Grain A railroad car moves at a constant speed of \(3.20 \mathrm{~m} / \mathrm{s}\) under a grain elevator. Grain drops into it at the rate of \(540 \mathrm{~kg} / \mathrm{min}\). What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

Short Answer

Expert verified
The force needed is \(28.8 \, \mathrm{N} \).

Step by step solution

01

- Identify given quantities

The railroad car has a constant speed of \(3.20 \, \mathrm{m/s}\) and grain drops at a rate of \(540 \, \mathrm{kg/min}\). Friction is negligible.
02

- Convert the grain dropping rate to SI units

Since the grain drops at \(540 \, \mathrm{kg/min}\), convert this to kilograms per second (kg/s). \( 540 \, \mathrm{kg/min} \, \times \, \left( \frac{1 \, \mathrm{min}}{60 \, \mathrm{s}} \right) = 9 \, \mathrm{kg/s} \)
03

- Use the momentum change principle

The force required to keep the railroad car moving at constant speed can be found from the change in momentum. The mass rate of grain falling is \( \frac{dm}{dt} = 9 \, \mathrm{kg/s} \). The rate of momentum change is: \( \frac{d(mv)}{dt} = v \frac{dm}{dt} \)
04

- Calculate the force

Substitute the given values into the equation: \( F = v \frac{dm}{dt} = 3.20 \, \mathrm{m/s} \, \times \, 9 \, \mathrm{kg/s} = 28.8 \, \mathrm{N} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

momentum change
Momentum is a fundamental concept in physics. It is defined as the product of an object's mass and velocity: \( p = mv \). When dealing with moving objects, especially those gaining or losing mass, it's crucial to understand how momentum changes over time.
In the given problem, we need to maintain a railroad car's momentum even as grain drops into it. The grain adds mass to the car, which affects the car's total momentum. The principle used here is the rate of momentum change, given by \( \frac{d(mv)}{dt} \). This tells us how the momentum is changing as mass flows into the car.
To find the required force, we use the fact that \( F = \frac{d(mv)}{dt} \) (force is the rate of change of momentum). By plugging in the velocity of the car and the rate at which mass (grain) is being added, we can calculate the force needed to keep the car at the same speed.
rate of mass flow
The rate of mass flow is another critical aspect when solving physics problems involving moving systems where mass is added or lost over time. In our problem, grain is added to the railroad car at a rate of \( 540 \, kg/min \). For ease and consistency, we first convert this to SI units (kilograms per second): \( \frac{540 \, kg}{60 \, s} = 9 \, kg/s \).
This rate tells us how quickly the car's mass is increasing every second. Understanding the rate of mass flow helps us to figure out how this addition affects the system's overall momentum and hence the force needed to keep things moving at a constant speed. Without converting and understanding this rate of flow, solving for the necessary force would be impossible.
constant speed
Constant speed is a simple yet essential aspect of the given problem. Speed is the rate at which an object covers distance. A constant speed means that the object moves at the same rate over time, without accelerating or decelerating. In our example, the railroad car moves with a constant speed of \( 3.2 \, m/s \).
Maintaining constant speed even as mass is added requires an understanding of Newton's first law of motion: an object in motion stays in motion with the same speed and in the same direction unless acted upon by unbalanced forces. When mass is added to the moving car, without a compensating force, the car would slow down. By calculating and applying the right force, \( 28.8 \, N \) in our problem, the car continues to move at constant speed, demonstrating practical application of physics principles.

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Most popular questions from this chapter

When jumping straight down, you can be seriously injured if you land stiff- legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. Suppose you have a mass \(m\) and you jump off a wall of height \(h\). (a) Use what you learned about constant acceleration motion to find the speed with which you hit the ground. Assume you simply step off the wall, so your initial \(y\) velocity is zero. Ignore air resistance. (Express your answer in terms of the symbols given.) (b) Suppose that the time interval starting when your feet first touch the ground until you stop is \(\Delta t .\) Calculate the (average) net force acting on you during that interval. (Again, express your answer in terms of the symbols given.) (c) Suppose \(h=1 \mathrm{~m}\). If you land stiff-legged, the time it takes you to stop may be as short as \(2 \mathrm{~ms}\), whereas if you bend your knees, it might be as long as \(0.1\) s. Calculate the average net force that would act on you in the two cases. (d) The net force on you while you are stopping includes both the force of gravity and the force of the ground pushing up. Which of these forces do you think does you the injury? Explain your reasoning. (e) For the two cases in part (c), calculate the upward force the ground exerts on you.

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