/*! 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 427 The mass of a car is \(1000 \mat... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 427

The mass of a car is \(1000 \mathrm{~kg}\). How much work is required to be done on it to make it move with a speed of \(36 \mathrm{~km} / \mathrm{h}\) ? (A) \(2.5 \times 10^{4} \mathrm{~J}\) (B) \(5 \times 10^{3} \mathrm{~J}\) (C) \(500 \mathrm{~J}\) (D) \(5 \times 10^{4} \mathrm{~J}\)

Short Answer

Expert verified
The work required to make the car move with a speed of \(36 \mathrm{~km} / \mathrm{h}\) is (D) \(5 \times 10^{4} \mathrm{~J}\).

Step by step solution

01

Convert the speed to m/s

First, we convert the speed from km/h to m/s by multiplying it by 1000 to convert kilometers to meters, then dividing by 3600 to convert hours to seconds. So, we have: \[ v = \frac{36 \text{ km/h} \times 1000 \text{m/km}}{3600 \text{s/h}} = 10 \text{ m/s} \]
02

Calculate the kinetic energy

Now that we have the speed in m/s, we can plug the mass (m = 1000 kg) and speed (v = 10 m/s) into the kinetic energy formula: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000\text{ kg} \times (10\text{ m/s})^2 \]
03

Compute the work done

Since work done is equal to the transfer of energy, and in this case, it is the transfer of kinetic energy, we can find the work done by solving the equation from step 2: \[ W = KE = \frac{1}{2} \times 1000\text{ kg} \times (10\text{ m/s})^2 = \frac{1}{2} \times 1000 \times 100 = 50,\!000\text{ J} \] From the given options, the work required to make the car move with a speed of 36 km/h is (D) \(5 \times 10^{4} \mathrm{~J}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When an object is in motion, it has the ability to cause change or do work due to its velocity and mass. The formula used to calculate kinetic energy is:
  • \( KE = \frac{1}{2} m v^2 \)
Where:
  • \( m \) is the mass of the object
  • \( v \) is the speed of the object

In the exercise, we calculated the kinetic energy of a car with a mass of 1000 kg moving at a speed of 10 m/s. By plugging these values into the kinetic energy formula, we find that the kinetic energy is 50,000 Joules. Understanding how to compute kinetic energy allows us to determine how much work is required to change an object's speed.
Conversion of Units
Often in physics, it's necessary to convert units to ensure calculations have a standard base unit system, typically the International System of Units (SI).
In our exercise, speed was given in kilometers per hour (km/h), which isn't suitable for most physics calculations, as they typically require meters per second (m/s).
  • To convert km/h to m/s, multiply the speed by 1000 to shift from kilometers to meters.
  • Then divide by 3600 to transform the hours into seconds.
Applying this conversion to the car's speed of 36 km/h, yields:
  • \( v = \frac{36 \times 1000}{3600} = 10 \text{ m/s} \)
Remember, accurate conversions ensure calculations are meaningful and follow accepted scientific conventions.
Work Done
Work done is intimately connected to energy transfer, and it's defined as the process of energy transfer when a force moves an object.
The formula for work when energy is converted from one form to another is:
  • \( W = KE \)
This relationship shows us that the work performed is equal to the change in kinetic energy. In the problem, the car's initial kinetic energy is zero since it's initially at rest. All calculated kinetic energy (50,000 Joules) is the work done to reach a speed of 36 km/h.

Knowing how work, energy, and force interrelate aids in designing systems where energy needs to be efficiently transformed and utilized. Understanding the concept of work done is essential in solving problems related to energy transfers in physics.

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Most popular questions from this chapter

If Wa, Wb, and Wc represent the work done in moving a particle from \(\mathrm{X}\) to \(\mathrm{Y}\) along three different path \(\mathrm{a}, \mathrm{b}\), and \(\mathrm{c}\) respectively (as shown) in the gravitational field of a point mass \(\mathrm{m}\), find the correct relation between \(\mathrm{Wa}, \mathrm{Wb}\) and \(\mathrm{Wc}\) (A) \(\mathrm{Wb}>\mathrm{Wa}>\mathrm{Wc}\) (B) \(\mathrm{Wa}<\mathrm{Wb}<\mathrm{Wc}\) (C) \(\mathrm{Wa}>\mathrm{Wb}>\mathrm{Wc}\) (D) \(\mathrm{Wa}=\mathrm{Wb}=\mathrm{Wc}\)

Two balls at same temperature collide. What is conserved (A) Temperature (B) velocity (C) kinetic energy (D) momentum

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)

An ice-cream has a marked value of \(700 \mathrm{kcal}\). How many kilo-watt- hour of energy will it deliver to the body as it is digested \((\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal})\) (A) \(0.81 \mathrm{kwh}\) (B) \(0.90 \mathrm{kwh}\) (C) \(1.11 \mathrm{kwh}\) (D) \(0.71 \mathrm{kwh}\)

Bansi does a given amount of work in \(30 \mathrm{sec}\). Jaimeen does the same amount of work .. in \(15 \mathrm{sec}\). The ratio of the output power of Bansi to the Jaimeen is.... (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(5: 3\)
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