/*! 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 can be found by first calculating the acceleration and the work done on the train and then using these values to calculate the average power using the formula P = W/t. The specific values can be obtained by substituting the given quantities into these formulas.

Step by step solution

01

Calculate the Acceleration

The train is accelerated from rest to 0.620 m/s in 21.0 ms. Using the formula for acceleration, which is final velocity minus initial velocity divided by the time taken, we get: \[a = \frac{{(v_f - v_i)}}{t}\]Substituting the given values:\[a = \frac{{0.620\, m/s - 0}}{21.0\, ms}\]After converting ms into s, \[a = \frac{{0.620\, m/s}}{21.0 \times 10^{-3} s}\]
02

Compute the Work Done

The work done by the motor on the train can be calculated using the work-energy theorem, which states that the work done is equal to the change in kinetic energy. The work done (W) is given by:\[W = \frac{1}{2} m a^2\]where m is the mass of the train. The mass, given as 875 g, should be converted to kg by dividing by 1000:\[m = \frac{{875\, g}}{1000} = 0.875\, kg\]Substitute the values of m and a in the above equation to find the Work Done.
03

Determine the Average Power

Power is defined as the work done per unit time. It can be represented as:\[P = \frac{W}{t}\]Substitute the value of work done (W) and time (t) to obtain the average power delivered to the train during its acceleration.

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object's velocity changes over time. It's a vector quantity, which means it has both a magnitude and a direction. In physics, it's often represented by the letter 'a' and calculated with the formula:
\[a = \frac{{(v_f - v_i)}}{t}\]
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time over which the change occurs. For the model train scenario, the train accelerated from rest, meaning its initial velocity was zero. With the formula and the given values, we calculated the train's acceleration during its 21.0 ms journey to reach the speed of 0.620 m/s. Understanding acceleration is crucial as it's a foundational concept in mechanics, influencing how we calculate the resultant forces and, as you'll see later, the kinetic energy.
Work-Energy Theorem
The work-energy theorem is a fundamental principle in physics that connects the concept of work with kinetic energy. It tells us that the work done on an object results in a change in its kinetic energy. The theorem is expressed as:
\[ W = \Delta KE \]
where W is work and \(\Delta KE\) is the change in kinetic energy. If an object starts from rest, the entire work done on it is converted into its kinetic energy. Applying the theorem to the problem at hand, we used it to determine the work performed by the electric motor of the model train. By equating the calculated work to the change in kinetic energy of the train, we laid the groundwork to later find the average power delivered during the train's acceleration.
Kinetic Energy
Kinetic energy is the energy of motion. An object possesses kinetic energy if it's moving, and the faster it moves, the greater its kinetic energy. The kinetic energy \(KE\) of an object can be calculated using the formula:
\[KE = \frac{1}{2}mv^2\]
where m is the mass and v is the velocity of the object. For the case of the model train, we considered its mass and the velocity it reached after being accelerated by the motor. Converting its mass from grams to kilograms was essential, as standard equations in physics typically require mass in kilograms. Once we knew the mass and the final speed, we were able to calculate the kinetic energy and from there, the work done by the motor on the train.
Physics Problem Solving
Physics problem solving is a critical skill that involves understanding the problem, identifying applicable concepts, and applying the appropriate formulas or principles. To solve physics problems efficiently, one should first analyze the given data, then follow a systematic approach: identify what you're solving for, choose the correct formula, substitute with the given values, and perform the calculation. In our exercise about the model train, we followed such a methodical approach. Starting from identifying acceleration, calculating the work done using the work-energy theorem, finding the kinetic energy, and finally determining the average power. Each step built upon the previous, demonstrating how physics problems can be broken down into a series of manageable tasks that lead to the solution.

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Most popular questions from this chapter

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by an engine is 175 hp, estimate the total frictional force acting on the car when it is moving at a speed of \(29 \mathrm{~m} / \mathrm{s}\). One horsepower cquals \(746 \mathrm{~W}\).

\(Q \mid C\) (a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-ofenergy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

An 80.0-kg skydiver jumps out of a balloon at an altitude of \(1000 \mathrm{~m}\) and opens the parachute at an altitude of \(200.0 \mathrm{~m}\). (a) Assuming that the total retarding force on the diver is constant at \(50.0 \mathrm{~N}\) with the parachute closed and constant at \(3600 \mathrm{~N}\) with the parachute open, what is the speed of the diver when he lands on the ground? (b) Do you think the skydiver will get hurt? Explain. (c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is \(5.00 \mathrm{~m} / \mathrm{s}\) ? (d) How realistic is the assumption that the total retarding force is constant? Explain. \(5.6\) Power

QIC The masses of the javelin, discus, and shot are \(0.80 \mathrm{~kg}, 2.0 \mathrm{~kg}\), and \(7.2 \mathrm{~kg}\), respectively, and record throws in the corresponding track events are about \(98 \mathrm{~m}, 74 \mathrm{~m}\), and \(23 \mathrm{~m}\), respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws, and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of \(2.0 \mathrm{~m}\). (c) Do your results suggest that air resistance is an important factor?

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway. 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?

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