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A train is traveling up a \(3.73^{\circ}\) incline at a speed of \(3.25 \mathrm{m} / \mathrm{s}\) when the last car breaks free and begins to coast without friction. (a) How long does it take for the last car to come to rest momentarily? (b) How far did the last car travel before mo-

Short Answer

Expert verified
(a) 5.10 seconds; (b) 8.30 meters

Step by step solution

01

Identify the Forces

When the last car of the train breaks free, it starts moving up an incline under the influence of gravity. Since there's no friction, the only force acting along the incline is the component of gravitational force. The component of gravitational force along the incline is given by \( mg\sin(\theta) \), where \( m \) is the mass of the car, \( g \) is the acceleration due to gravity \( (9.8 \text{ m/s}^2) \), and \( \theta \) is the incline angle \(3.73^\circ\).
02

Calculate the Gravitational Acceleration

The acceleration due to gravity along the incline \( a \) can be found using:\[ a = g \sin(\theta) = 9.8 \times \sin(3.73^\circ) \]We need to find \( \sin(3.73^\circ) \) using a calculator. This yields approximately 0.065. Thus:\[ a \approx 9.8 \times 0.065 \approx 0.637 \text{ m/s}^2 \]
03

Use Kinematic Equation for Time to Rest

We use the kinematic equation \( v = u + at \), where \( v = 0 \) m/s (final velocity), \( u = 3.25 \) m/s (initial velocity), and \( a = -0.637 \text{ m/s}^2 \) (negative because it's deceleration). Solving for time \( t \):\[ 0 = 3.25 - 0.637t \]\[ 0.637t = 3.25 \]\[ t = \frac{3.25}{0.637} \approx 5.10 \text{ seconds} \]
04

Use Kinematic Equation for Distance

Now, to find the distance traveled, we use the equation:\[ s = ut + \frac{1}{2}at^2 \]Plugging the values \( u = 3.25 \) m/s, \( a = -0.637 \text{ m/s}^2 \), and \( t = 5.10 \) s:\[ s = 3.25 \times 5.10 + \frac{1}{2} \times (-0.637) \times (5.10)^2 \]\[ s = 16.58 - 8.28 \approx 8.30 \text{ meters} \]

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Key Concepts

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

Gravitational Force
Gravitational force is an essential concept in kinematics, particularly when dealing with objects moving on an inclined plane. It is the force that attracts a body towards the center of the Earth. This force is what keeps us grounded and is constantly acting on every object with mass. The magnitude of the gravitational force can be calculated using the formula:
  • \[ F = mg \]
where\( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).
In the case of an inclined plane, the gravitational force can be split into two components:
  • The normal force, perpendicular to the plane
  • The parallel force, along the plane, causing motion
The parallel component that causes objects to slide down the slope is crucial here, and it can be defined by \( mg\sin(\theta) \), where \( \theta \) is the angle of the incline. This component is what affects the acceleration of the train car as it moves up the incline.
Inclined Plane
An inclined plane is simply a flat surface tilted at an angle, rather than being horizontal. It is one of the classic simple machines used to ease the lifting of a heavy object by spreading the required effort over a greater distance. However, in kinematics, an inclined plane affects how objects move under the influence of gravity.
Since the plane is inclined, objects tend to slide down due to the gravitational force's component, as mentioned earlier. For instance, the train car in our example was subject to this very force after breaking free from the rest of the train.
  • When analyzing motion on an incline, the angle \( \theta \) is important. It determines how steep the incline is.
  • The steeper the incline, the greater the gravitational component acting along it and the faster objects will accelerate down the slope.

When the last car of the train begins to coast along the inclined plane without friction, it is primarily moving due to gravity's pull. The incline angle of \( 3.73^\circ \) significantly affects how quickly it slows and stops.
Kinematic Equations
Kinematic equations are fundamental in physics for predicting the future motion of objects. They relate the five key aspects of motion: displacement, initial velocity, final velocity, acceleration, and time.
The main equations include:
  • \[ v = u + at \]
  • \[ s = ut + \frac{1}{2}at^2 \]
  • \[ v^2 = u^2 + 2as \]
These equations are invaluable for solving problems like when the last car of a train stops on an incline. In our example, two of these equations helped determine how long it took for the car to rest momentarily, and the distance it traveled during this time.
By knowing the initial velocity \( u \) and recognizing that the final velocity \( v \) is zero (since it comes to a stop), you can calculate how time and distance are affected by the negative acceleration \( a \) due to the gravitational component. These equations provide a systematic approach to solving real-world problems involving motion.

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Most popular questions from this chapter

When you lift a bowling ball with a force of \(82 \mathrm{N}\), the ball accelerates upward with an acceleration a. If you lift with a force of \(92 \mathrm{N}\), the ball's acceleration is \(2 \mathrm{a}\). Find (a) the weight of the bowling ball, and (b) the acceleration \(a\).

" IP A drag racer crosses the finish line doing \(202 \mathrm{mi} / \mathrm{h}\) and promptly deploys her drag chute (the small parachute used for braking). (a) What force must the drag chute exert on the \(891-\mathrm{kg}\) ear to slow it to \(45.0 \mathrm{mi} / \mathrm{h}\) in a distance of \(185 \mathrm{m} ?\) (b) Describe the strategy you used to solve part (a).

A \(0.53-\mathrm{kg}\) billiard ball initially at rest is given a speed of \(12 \mathrm{m} / \mathrm{s}\) during a time interval of \(4.0 \mathrm{ms}\). What average force acted on the ball during this time?

A driver who does not wear a seatbelt continues to move forward with a speed of \(18.0 \mathrm{m} / \mathrm{s}\) (due to inertia) until something solid like the steering wheel is encountered. The driver now comes to rest in a much shorter distance - perhaps only a few centimeters. Find the magnitude of the net force acting on a \(65.0-\mathrm{kg}\) driver who is decelerated from \(18.0 \mathrm{m} / \mathrm{s}\) to rest in \(5.00 \mathrm{cm} .\) A. \(3240 \mathrm{N}\) B. \(1.17 \times 10^{4} \mathrm{N}\) C. \(2.11 \times 10^{5} \mathrm{N}\) D. \(4.21 \times 10^{5} \mathrm{N}\)

\(\cdot\) Suppose a rocket launches with an acceleration of \(30.5 \mathrm{m} / \mathrm{s}^{2}\) What is the apparent weight of an \(92-\mathrm{kg}\) astronaut aboard this rocket?

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