/*! 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 53 The electric motor of a model tr... [FREE SOLUTION] | 91Ó°ÊÓ

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The electric motor of a model train accelerates the train from rest to \(0.620 \mathrm{~m} / \mathrm{s}\) in \(21.0 \mathrm{~ms}\), The total mass of the train is \(875 \mathrm{~g} .\) Find the average power delivered to the train during its acceleration.

Short Answer

Expert verified
The average power delivered to the train during its acceleration is 26 Watts.

Step by step solution

01

Find the Acceleration

Since the train starts from rest and accelerates to a final speed, we can use the kinematic formula \(a = \frac{v - u}{t}\) to find the acceleration. Here, \(v = 0.620 m/s\) is the final speed, \(u = 0 m/s\) is the initial speed (since the train starts from rest), and \(t = 21.0 ms = 0.021 s\) is the time of acceleration. Substituting these values, we get \(a = \frac{0.620 - 0}{0.021} = 29.52 m/s^2\).
02

Calculate the Work Done

We know that work done is the product of force and displacement. In this case, we consider force to be the product of mass and acceleration (from Newton's second law, \(F = ma\)), and displacement to be the product of speed and time. The mass is given as \(875 g = 0.875 kg\). Substituting the values, we get \(W = Ft = mat = 0.875 * 29.52 * 0.021 = 0.546 Joules\).
03

Calculate the Average Power

Now, the average power can be found by the equation \(P_{avg} = \frac{W}{t}\), where \(W\) is work done and \(t\) is time during which the work is done. Substituting the obtained values, we get \(P_{avg} = \frac{0.546}{0.021} = 26 Watts\).

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Key Concepts

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

Kinematic Formulas
Kinematic formulas are essential tools for describing the motion of objects. They relate the variables of an object's movement: velocity, acceleration, time, and displacement. One fundamental formula is the equation for acceleration
, which is based on the change in velocity over time. In the exercise, the train’s acceleration (
\tagain\textual\tagain) was found using the formula
\(a = \frac{v - u}{t}\)
, where
\tagain\textual\tagain\(v\)
is final velocity,
\tagain\textual\tagain\(u\)
is initial velocity, and
\tagain\textual\tagain\(t\)
is the time taken to change velocity. These formulas are crucial for predicting an object’s future position or determining its past movement, making them a cornerstone in physics problems involving motion.
Newton's Second Law
Newton's second law of motion provides a quantitative description of the changes in motion that a force can produce. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass
\(a = \frac{F}{m}\)
. This relationship is fundamental in calculating the force in the problem. The law reminds us that to propel an object forward, such as our model train, a certain amount of force must be applied. The stronger the force or the lighter the object, the greater the acceleration. In our case, we calculated the train's force by rearranging Newton's second law to
\(F = ma\)
and then used that force to find the work done.
Work-Energy Principle
The work-energy principle is a concept that bridges the gap between force and energy. It states that the work done by forces on an object results in a change in the object’s kinetic energy. Work (
\(W\)
) is calculated as the force applied to an object times the distance over which that force is applied, or
\(W = F * d\)
. In the context of our problem, we use the work done by the electric motor's force to accelerate the train. As force was derived using Newton's second law, the displacement can be estimated from kinematic equations. Connecting these concepts, we then used the total work to arrive at the average power, which is the rate at which the train's kinetic energy changed over time.
Acceleration
Acceleration is the rate at which an object’s velocity changes with time. It is a vector quantity, having both a magnitude and a direction. In the example we've been discussing, the model train accelerated from rest, meaning it had an initial velocity of zero. The exercise utilized the straightforward formula for acceleration when only the velocities and time are known. This concept is pivotal because acceleration is a result of forces acting on an object (as per Newton's second law) and is intimately linked with the object's kinetic energy changes, providing a gateway to understanding the work-energy principle. In our daily life, acceleration is experienced whenever we start moving from a stop, speed up, slow down, or change direction.

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Most popular questions from this chapter

A \(5.75-\mathrm{kg}\) object is initially moving so that its \(x\) -component of velocity is \(6.00 \mathrm{~m} / \mathrm{s}\) and its \(y\) -component of velocity is \(-2.00 \mathrm{~m} / \mathrm{s}\), (a) What is the kinetic energy of the object at this time? (b) Find the change in kinetic energy of the object if its velocity changes so that its new \(x\) -component is \(8.50 \mathrm{~m} / \mathrm{s}\) and its new \(y\) -component is \(5.00 \mathrm{~m} / \mathrm{s}\).

The force acting on an object is given by \(F_{n}=(8 x-16) \mathrm{N}\), where \(x\) is in meters. (a) Make a plot of this force versus \(x\) from \(x=0\) to \(x=3.00 \mathrm{~m} .\) (b) From your graph, find the net work done by the force as the object moves from \(x=0\) to \(x=3.00 \mathrm{~m}\).

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway (Fig. P5.83). He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point \(10.0 \mathrm{~m}\) above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a \(5.00-\mathrm{m}\) length of the cord, the jumper finds that his body weight stretches it by \(1.50 \mathrm{~m} .\) He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?

A \(7.00-\mathrm{kg}\) bowling ball moves at \(3.00 \mathrm{~m} / \mathrm{s}\). How fast must a \(2.45\) -g Ping-Pong ball move so that the two balls have the same kinetic energy?

A ski jumper starts from rest \(50.0 \mathrm{~m}\) above the ground on a frictionless track and flies off the uack at an angle of \(45.0^{\circ}\) above the horizontal and at a height of \(10.0 \mathrm{~m}\) above the level ground. Neglect air resistance. (a) What is her speed when she leaves the track? (b) What is the maximum altitude she attains after leaving the track? (c) Where does she land relative to the end of the track?

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