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Chapter 2: Problem 6

An object of mass \(1000 \mathrm{~kg}\), initially having a velocity of \(100 \mathrm{~m} / \mathrm{s}\), decelerates to a final velocity of \(20 \mathrm{~m} / \mathrm{s}\). What is the change in kinetic energy of the object, in \(\mathrm{kJ}\) ?

Short Answer

Expert verified
The change in kinetic energy is 4800 kJ.

Step by step solution

01

Recall the Kinetic Energy Formula

The kinetic energy (KE) of an object is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
02

Calculate Initial Kinetic Energy

Use the initial velocity (\( v_i = 100 \mathrm{~m/s} \)) and the mass (\( m = 1000 \mathrm{~kg} \)) to find the initial kinetic energy. \[ KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2} \times 1000 \mathrm{~kg} \times (100 \mathrm{~m/s})^2 = 5000000 \mathrm{~J} = 5000 \mathrm{~kJ} \]
03

Calculate Final Kinetic Energy

Use the final velocity (\( v_f = 20 \mathrm{~m/s} \)) and the mass (\( m = 1000 \mathrm{~kg} \)) to find the final kinetic energy. \[ KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2} \times 1000 \mathrm{~kg} \times (20 \mathrm{~m/s})^2 = 200000 \mathrm{~J} = 200 \mathrm{~kJ} \]
04

Determine the Change in Kinetic Energy

Subtract the final kinetic energy from the initial kinetic energy to find the change in kinetic energy. \[ \Delta KE = KE_i - KE_f = 5000 \mathrm{~kJ} - 200 \mathrm{~kJ} = 4800 \mathrm{~kJ} \]

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Key Concepts

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

kinetic energy formula
Kinetic energy (KE) is the energy possessed by an object due to its motion. It's an important concept in physics that helps us understand how energy is transferred and transformed in moving objects. The formula for calculating kinetic energy is: \[ KE = \frac{1}{2}mv^2 \]Here,
  • \(m\) is the mass of the object.
  • \(v\) is the velocity of the object.
When we use this formula, we can determine how much energy an object has because of its speed and mass. It's crucial to remember that velocity is squared in this formula, which means that even small changes in velocity will have a big impact on kinetic energy.
initial kinetic energy
The initial kinetic energy of an object is the amount of kinetic energy it possesses before any changes in speed or direction. Using the initial velocity and the mass of the object, we can find out this energy.
For example, in the given exercise:
  • The object has a mass \(m = 1000 \mathrm{~kg}\).
  • The initial velocity \(v_i = 100 \mathrm{~m/s}\).

Using the kinetic energy formula, the initial kinetic energy is:

\[KE_i = \frac{1}{2}mv_i^2\]
Plugging in our values, we get:

\[KE_i = \frac{1}{2} \times 1000 \mathrm{~kg} \times (100 \mathrm{~m/s})^2 \]After calculation, it results in:

\[KE_i = 5000000 \mathrm{~J} = 5000 \mathrm{~kJ}\]
This large amount of energy shows how much energy the object has due to its high initial speed.
final kinetic energy
The final kinetic energy is the amount of kinetic energy the object has after it has decelerated or changed its velocity.
In the exercise, the object slows down to a final velocity of:
  • \(v_f = 20 \mathrm{~m/s}\)

We use the same kinetic energy formula but with the final velocity:
\[KE_f = \frac{1}{2}mv_f^2\]
Substitute the known values:
\[KE_f = \frac{1}{2} \times 1000 \mathrm{~kg} \times (20 \mathrm{~m/s})^2\]
This calculates to:
\[KE_f = 200000 \mathrm{~J} = 200 \mathrm{~kJ}\]
The significant decrease in kinetic energy indicates how much energy was lost due to the reduction in speed.
change in kinetic energy
The change in kinetic energy tells us how much kinetic energy has been lost or gained by the object as it changes its velocity. This is found by subtracting the final kinetic energy from the initial kinetic energy.
In the exercise, we calculated:
  • Initial kinetic energy \(KE_i = 5000 \mathrm{~kJ}\).
  • Final kinetic energy \(KE_f = 200 \mathrm{~kJ}\).

The change in kinetic energy (\(\Delta KE\)) is given by:
\[\Delta KE = KE_i - KE_f\]
Substitute the values and solve:
\[\Delta KE = 5000 \mathrm{~kJ} - 200 \mathrm{~kJ}\]
So,\(\Delta KE = 4800 \mathrm{~kJ}\)
This means the object has lost 4800 kilojoules of kinetic energy as it slowed down from 100 m/s to 20 m/s.

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Most popular questions from this chapter

A gas contained within a piston-cylinder assembly undergoes two processes, \(A\) and \(B\), between the same end states, 1 and 2 , where \(p_{1}=10\) bar, \(V_{1}=0.1 \mathrm{~m}^{3}, U_{1}=400 \mathrm{~kJ}\) and \(p_{2}=1\) bar, \(V_{2}=1.0 \mathrm{~m}^{3}, U_{2}=200 \mathrm{~kJ}\) : Process A: Process from 1 to 2 during which the pressurevolume relation is \(p V=\) constant. Process B: Constant-volume process from state 1 to a pressure of 2 bar, followed by a linear pressure-volume process to state 2 . Kinetic and potential energy effects can be ignored. For each of the processes \(A\) and \(B\), (a) sketch the process on \(p-V\) coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

An automobile weighing \(2500 \mathrm{lbf}\) increases its gravitational potential energy by \(2.25 \times 10^{4}\) Btu in going from an elevation of \(5183 \mathrm{ft}\) in Denver to the highest elevation on Trail Ridge Road in the Rocky Mountains. What is the elevation at the high point of the road, in \(\mathrm{ft}\) ?

A construction crane weighing \(12,000 \mathrm{lbf}\) fell from a height of \(400 \mathrm{ft}\) to the street below during a severe storm. For \(g=\) \(32.05 \mathrm{ft} / \mathrm{s}^{2}\), determine the mass, in \(\mathrm{lb}\), and the change in gravitational potential energy of the crane, in \(\mathrm{ft} \cdot \mathrm{lbf}\).

A gas contained in a piston-cylinder assembly undergoes two processes, \(\mathrm{A}\) and \(\mathrm{B}\), between the same end states, 1 and 2 , where \(p_{1}=1\) bar, \(V_{1}-1 \mathrm{~m}^{3}, U_{1}=400 \mathrm{~kJ}\) and \(p_{2}-10\) bar, \(V_{2}=0.1 \mathrm{~m}^{3}, U_{2}=450 \mathrm{~kJ}\) : Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constant-pressure process to state \(2 .\) Process B: Process from 1 to 2 during which the pressurevolume relation is \(p V=\) constant. Kinetic and potential effects can be ignored. For each of the processes \(\mathrm{A}\) and \(\mathrm{B}\), (a) sketch the process on \(p-V\) coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in. \(^{2}\) and weight of \(100 \mathrm{lbf}\). The atmosphere exerts a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by \(1 \mathrm{ft}\). The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.
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