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A railroad car moves under a grain elevator at a constant speed of \(3.20 \mathrm{~m} / \mathrm{s}\). Grain drops into the car 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 required force is 28.8 N.

Step by step solution

01

Convert Grain Rate to SI Units

First, we need to convert the rate at which grain drops into the car from kg/min to kg/s (since SI units use seconds). Use the conversion: \[ 540 \frac{\text{kg}}{\text{min}} = \frac{540}{60} \frac{\text{kg}}{\text{s}} = 9 \frac{\text{kg}}{\text{s}} \]
02

Identify Forces and Apply Newton's Second Law

Since the car is moving at a constant speed and friction is negligible, the only force we need to consider is the force required to keep the additional mass moving at the same velocity. According to Newton's Second Law, the force can be calculated as \( F = \Delta p / \Delta t \), where \( p \) is momentum and \( t \) is time.
03

Calculate Change in Momentum

Momentum \( p \) is given by \( p = mv \). The grain's mass change per second is \( 9 \text{ kg/s} \), moving at \( 3.20 \text{ m/s} \). Thus, the change in momentum is: \[ \Delta p = 9 \times 3.20 = 28.8 \text{ kg} \cdot \text{m/s} \] per second.
04

Determine Required Force

Using the relation \( F = \Delta p / \Delta t \), where \( \Delta p = 28.8 \text{ kg} \cdot \text{m/s} \) and \( \Delta t = 1 \text{ s} \), the required force is: \[ F = 28.8 \text{ N} \]

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Key Concepts

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

Momentum
Momentum is a measure of the motion of an object and is a fundamental concept in physics. It helps us understand how objects interact when they collide or exert forces on each other. The momentum of an object is calculated by multiplying its mass by its velocity, given by the formula \(p = mv\).
  • Here, \(p\) represents momentum.
  • \(m\) is the mass of the object.
  • \(v\) is the velocity of the object.
In our exercise, the grain adds onto the moving railroad car, changing its overall momentum. Although the car moves at a constant speed, the continuous mass being added from the grain requires additional force to maintain this speed. The change in momentum in this context is \(p = 9 \text{ kg/s} \times 3.20 \text{ m/s} = 28.8 \text{ kg} \cdot \text{m/s} \). This change represents how much momentum is gained per second as a result of the grain falling into the car.
SI Units Conversion
Units serve as a universally understood language in science, allowing scientists worldwide to understand and communicate findings accurately. The International System of Units (SI) is the most widely recognized system of units. In any physics problem, it’s crucial to convert given values into SI units before calculations. For expressing rates in SI units, if a rate is provided per minute, that's denoted as \( \text{min}^{-1} \), it must be converted to per second, or \( \text{s}^{-1} \), since seconds are the base unit of time in SI.
In our example, the rate of grain dropping is initially given as \(540 \text{ kg/min}\). To convert this to kilograms per second, divide by 60 because there are 60 seconds in a minute:\[540 \frac{\text{kg}}{\text{min}} = \frac{540}{60} \frac{\text{kg}}{\text{s}} = 9 \frac{\text{kg}}{\text{s}}\] This conversion ensures that the rate is suitable for further physics calculations involving momentum and force.
Force Calculation
Newton's Second Law of Motion states that the force acting on an object is equal to the rate of change of its momentum. In formula terms, it is expressed as \(F = \frac{\Delta p}{\Delta t}\). This means the force is calculated based on how much the momentum changes over time.
In our exercise, the grain adds mass to the railroad car without changing its constant velocity. This results in a continued increase in momentum that requires force to sustain the car's motion at constant velocity. We calculated the momentum change as \(\Delta p = 28.8 \text{ kg} \cdot \text{m/s}\).Since this change happens every second, we set \(\Delta t = 1\ \text{s}\) and thus derive the force to maintain constant speed:\[F = \frac{\Delta p}{\Delta t} = \frac{28.8 \text{ kg} \cdot \text{m/s}}{1 \text{ s}} = 28.8 \text{ N}\] This means 28.8 Newtons of force are required to keep the car uninterrupted in its motion as the grain continues to fall.

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Most popular questions from this chapter

In February 1955 , a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

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