/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 An \(8000-\mathrm{kg}\) engine p... [FREE SOLUTION] | 91Ó°ÊÓ

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An \(8000-\mathrm{kg}\) engine pulls a 40000 -kg train along a level track and gives it an acceleration \(a_{1}=1.20 \mathrm{~m} / \mathrm{s}^{2}\). What acceleration \(\left(a_{2}\right)\) would the engine give to a \(16000-\mathrm{kg}\) train? For a given engine force, the acceleration is inversely proportional to the total mass. Thus, $$ a_{2}=\frac{m_{1}}{m_{2}} a_{1}=\frac{8000 \mathrm{~kg}+40000 \mathrm{~kg}}{8000 \mathrm{~kg}+16000 \mathrm{~kg}}\left(1.20 \mathrm{~m} / \mathrm{s}^{2}\right)=2.40 \mathrm{~m} / \mathrm{s}^{2} $$

Short Answer

Expert verified
The acceleration for the 16000 kg train is 2.40 m/s².

Step by step solution

01

Understand the relationship

The acceleration of the train is inversely proportional to its total mass. This means that if the engine force is constant, a larger mass results in a smaller acceleration, and vice versa.
02

Identify given values

For the first scenario, the mass of the engine is 8000 kg and the mass of the train is 40000 kg. The given acceleration (\(a_1\)) is 1.20 m/s². For the second scenario, the train's mass changes to 16000 kg.
03

Calculate total mass for each scenario

First scenario: total mass \(m_1 = 8000 \,\text{kg} + 40000 \,\text{kg} = 48000 \,\text{kg}\). Second scenario: total mass \(m_2 = 8000 \,\text{kg} + 16000 \,\text{kg} = 24000 \,\text{kg}\).
04

Apply the formula to find new acceleration

Use the relation \(a_2 = \frac{m_1}{m_2} a_1\) to find the acceleration for the second train. Substitute the values: \(a_2 = \frac{48000}{24000} \times 1.20\), simplifying to \(a_2 = 2 \times 1.20 = 2.40\,\text{m/s}^2\).

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Key Concepts

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

Force and acceleration
In physics, force is directly related to the acceleration of an object. When a force is applied to an object, it can change the object's speed or direction. This change in motion is called acceleration. Imagine pushing a toy car; the harder you push, the faster it moves. This demonstrates how force affects acceleration.

Acceleration itself is measured in meters per second squared (m/s²), indicating how fast the velocity of an object changes over time. Without any force, an object remains at rest or continues at constant velocity. This is a fundamental principle of motion.

If the same force is applied to different objects, the object with a smaller mass will accelerate more than one with a larger mass, if friction and other forces are negligible. This relationship between force and acceleration is not only intuitive but also mathematically described in Newton's laws of motion.
Newton's second law
Newton's second law of motion is a cornerstone in understanding how forces interact with masses. The law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.\[F = ma\]This equation means that if you know any two of the values (force, mass, or acceleration), you can find the third.
  • Force: It is the push or pull exerted on an object, measured in Newtons (N).
  • Mass: Represents how much matter an object contains, measured in kilograms (kg).
  • Acceleration: The rate of change of velocity, as mentioned earlier, measured in meters per second squared (m/s²).
Newton's second law helps us understand why heavier objects require more force to accelerate at the same rate as lighter ones. It is essential for solving problems involving the motion of objects, just like in the train scenario.
Mass and acceleration relationship
Understanding the relationship between mass and acceleration is crucial in physics. The two are inversely proportional when the force is constant. This concept means that as the mass of an object increases, its acceleration decreases, provided the force applied remains the same.For example, if an engine pulls a train, the acceleration achieved depends not just on the engine's force but also on the mass of the train. In our exercise, as the train's mass decreases from 40000 kg to 16000 kg, the acceleration increases. This relationship is mathematically represented by rearranging Newton's second law:\[a = \frac{F}{m}\]This formula shows that the acceleration (a) is calculated by dividing the force (F) by the mass (m). So, for a constant force, a larger mass results in smaller acceleration, and vice versa.
Proportional reasoning
Proportional reasoning is a handy tool in solving physics problems. It allows you to understand how changes in one quantity affect another. In simpler terms, it's about seeing how different variables relate to each other.In our train exercise, we used proportional reasoning to predict how the change in mass affects acceleration. By understanding that if the mass is halved, the acceleration doubles for the same force, we can calculate without knowing the exact force applied.Such reasoning involves directly using ratios. In the exercise:\[a_2 = \frac{m_1}{m_2} a_1\]Here, the ratio of the masses (\frac{m_1}{m_2}) directly modified the initial acceleration (a_1) to give the new acceleration (a_2). Being comfortable with proportional reasoning helps solve many physics problems more intuitively and efficiently.

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Most popular questions from this chapter

Imagine a planet having a mass twice that of Earth and a radius equal to \(1.414\) times that of Earth. Determine the acceleration due to gravity at its surface.

A car coasting at \(20 \mathrm{~m} / \mathrm{s}\) along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coefficient between tires and road is \(0.90 ?\) Assume that all four wheels brake identically. If the brakes don't lock, the car stops via static friction. The friction force at one wheel, call it wheel 1 , is $$ F_{\mathrm{fl}}=\mu_{5} F_{N 1}=\mu_{s} F_{W 1} $$ where \(F_{W}\) is the weight carried by wheel 1 . We obtain the total friction force \(F_{\mathrm{f}}\) by adding such terms for all four wheels: $$ F_{\mathrm{f}}=\mu_{s} F_{W 1}+\mu_{s} F_{W_{2}}+\mu_{s} F_{W 3}+\mu_{s} F_{W 4}=\mu_{s}\left(F_{W_{1}}+F_{W 2}+F_{W 3}+F_{W_{4}}\right)=\mu_{s} F_{w} $$ where \(F_{W}\) is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This friction force is the only unbalanced force on the car (we neglect air friction). Writing \(F=m a\) for the car with \(F\) replaced by \(-\mu_{s} F_{W}\) gives \(-\mu_{s} F_{W}=m a\), where \(m\) is the car's mass and the positive direction is taken as the direction of motion. However, \(F_{W}=m g\); so the car's acceleration is $$ a=-\frac{\mu_{s} F_{W}}{m}=-\frac{\mu_{s} m g}{m}=-\mu_{s} g=(-0.90)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=-8.829 \mathrm{~m} / \mathrm{s}^{2} $$ We can determine how far the car went before stopping by solving the constant- \(a\) motion problem. Knowing that \(v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-8.829 \mathrm{~m} / \mathrm{s}^{2}\), we find from \(v_{f}^{2}-v_{i}^{2}=2 a x\) that $$ x=\frac{(0-400) \mathrm{m}^{2} / \mathrm{s}^{2}}{-17.66 \mathrm{~m} / \mathrm{s}^{2}}=22.65 \mathrm{~m} \quad \text { or } \quad 23 \mathrm{~m} $$ If the four wheels had not all been braking optimally, the stopping distance would have been longer.

Just as her parachute opens, a \(60-\mathrm{kg}\) parachutist is falling at a speed of \(50 \mathrm{~m} / \mathrm{s}\). After \(0.80 \mathrm{~s}\) has passed, the chute is fully open and her speed has dropped to \(12.0 \mathrm{~m} / \mathrm{s}\). Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform.

Typically, a bullet leaves a standard 45 -caliber pistol (5.0-in. barrel) at a speed of \(262 \mathrm{~m} / \mathrm{s}\). If it takes \(1 \mathrm{~ms}\) to traverse the barrel, determine the average acceleration experienced by the \(16.2-\mathrm{g}\) bullet within the gun, and then compute the average force exerted on it.

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

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