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A \(1500 \mathrm{~kg}\) car starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is its kinetic energy at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the 30 s interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?

Short Answer

Expert verified
(a) 300,000 J; (b) 10,000 W; (c) 20,000 W.

Step by step solution

01

Convert Speed to Meters per Second

The initial speed of the car is given in kilometers per hour (km/h), so we need to convert it to meters per second (m/s) to be consistent with the SI unit system. Use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). Thus, \(72 \text{ km/h} = \frac{72}{3.6} \text{ m/s} = 20 \text{ m/s}\).
02

Calculate Final Kinetic Energy

Kinetic energy (KE) is given by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the final velocity. Substituting the given values, \( KE = \frac{1}{2} \times 1500 \text{ kg} \times (20 \text{ m/s})^2 = 300,000 \text{ J} \).
03

Determine Average Power

Average power is the total work done over time, which can be calculated using the formula \( P_{avg} = \frac{W}{t} \). Here, the work done \(W\) is equal to the change in kinetic energy (as the car started from rest). Thus, \( P_{avg} = \frac{300,000 \text{ J}}{30 \text{ s}} = 10,000 \text{ W}\).
04

Calculate Instantaneous Power at the End of 30 Seconds

Instantaneous power at any point under constant acceleration can be calculated using \(P = F \cdot v\), where \(F\) is the force and \(v\) is the velocity at that point. Since acceleration is constant: \(F = m \cdot a\). The acceleration \(a = \frac{\Delta v}{\Delta t} = \frac{20 \text{ m/s}}{30 \text{ s}} = \frac{2}{3} \text{ m/s}^2\). Therefore, \(F = 1500 \text{ kg} \times \frac{2}{3} \text{ m/s}^2 = 1000 \text{ N} \). Then, \( P = 1000 \text{ N} \times 20 \text{ m/s} = 20,000 \text{ W}\).

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Key Concepts

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

Power in Physics: A Practical Insight
Power is an essential concept in physics that refers to the rate at which work is done or energy is transferred. In practical terms, power tells us how quickly a car's engine can perform the work needed to increase the car's speed.
For example, in our exercise, we calculate average power to understand the energy needed by the car engine over a 30-second interval. Average power is calculated using the formula \( P_{avg} = \frac{W}{t} \), where \(W\) is the work done, which in this case, corresponds to the car’s increase in kinetic energy.
To find the instantaneous power at the end of the 30 seconds, we use the formula \(P = F \cdot v\). Here, force \(F\) is derived from the product of mass \(m\) and constant acceleration \(a\), and \(v\) is the velocity at that point. This tells us how much energy the car's engine needs to maintain or increase the speed at that instant.
Understanding Acceleration in Motion
Acceleration is a key concept when dealing with moving objects. It defines the rate of change of velocity of an object. In our exercise, we assume constant acceleration to simplify calculations.
Constant acceleration means the car's speed increases uniformly over time. We calculate acceleration using the formula \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time over which this change occurs.
This concept is crucial when calculating other aspects, like instantaneous power, because force exerted on the car is directly proportional to acceleration (\(F = m \cdot a\)). Understanding this connection helps us determine how different forces impact a car's motion.
The Work-Energy Principle Explained
The work-energy principle is an important aspect of physics that states that the work done on an object is equal to the change in its kinetic energy. In simple terms, it means that any energy transferred into doing work on an object transforms into kinetic energy.
In our exercise, the car starts from rest, so its initial kinetic energy is zero. The engine does work to increase the car’s speed, and this work is precisely what is needed to reach the final kinetic energy. The formula for kinetic energy \( KE = \frac{1}{2} m v^2 \) allows us to determine how much energy is involved.
Understanding the work-energy principle is vital for solving problems related to energy transfer, making it easier to comprehend how forces and movements are interconnected in physical systems.

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Most popular questions from this chapter

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m} .\) The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0 \mathrm{~cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. If your friend's outstretched hands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. \(8-30\) ), (a) how much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U(\mathrm{c})\) when the book is released and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U,(\mathrm{~g}) U\) at the release point, and (h) \(U\) at her hands.

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

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