What is the work that needs to be done to increase the speed of a 1 kg ball from 2 m/s to 4 m/s?
6 J
Understanding Work and Energy in Physics
The question asks for the work required to change the speed of an object. In physics, the work done on an object is often related to its change in kinetic energy. This relationship is described by the Work-Energy Theorem.
Applying the Work-Energy Theorem
The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy. Mathematically, this is written as:
\(W = \Delta KE = KE_{final} - KE_{initial}\)
Where:
- \(W\) is the work done on the object.
- \(\Delta KE\) is the change in kinetic energy.
- \(KE_{initial}\) is the initial kinetic energy.
- \(KE_{final}\) is the final kinetic energy.
The kinetic energy of an object with mass \(m\) and speed \(v\) is given by:
\(KE = \frac{1}{2}mv^2\)
Step-by-Step Calculation of Work Done
We are given the following information:
- Mass of the ball, \(m = 1 \text{ kg}\)
- Initial speed of the ball, \(v_i = 2 \text{ m/s}\)
- Final speed of the ball, \(v_f = 4 \text{ m/s}\)
1. Calculate the Initial Kinetic Energy (\(KE_{initial}\)):
Using the formula \(KE = \frac{1}{2}mv^2\):
\(KE_{initial} = \frac{1}{2} \times m \times v_i^2\)
\(KE_{initial} = \frac{1}{2} \times 1 \text{ kg} \times (2 \text{ m/s})^2\)
\(KE_{initial} = \frac{1}{2} \times 1 \times 4 \text{ J}\)
\(KE_{initial} = 2 \text{ J}\)
2. Calculate the Final Kinetic Energy (\(KE_{final}\)):
Using the formula \(KE = \frac{1}{2}mv^2\):
\(KE_{final} = \frac{1}{2} \times m \times v_f^2\)
\(KE_{final} = \frac{1}{2} \times 1 \text{ kg} \times (4 \text{ m/s})^2\)
\(KE_{final} = \frac{1}{2} \times 1 \times 16 \text{ J}\)
\(KE_{final} = 8 \text{ J}\)
3. Calculate the Work Done (\(W\)):
Using the Work-Energy Theorem \(W = KE_{final} - KE_{initial}\):
\(W = 8 \text{ J} - 2 \text{ J}\)
\(W = 6 \text{ J}\)
Therefore, the work that needs to be done to increase the speed of the 1 kg ball from 2 m/s to 4 m/s is 6 J.
| State | Speed (\(v\)) | Mass (\(m\)) | Kinetic Energy (\(KE = \frac{1}{2}mv^2\)) |
|---|---|---|---|
| Initial | \(2 \text{ m/s}\) | \(1 \text{ kg}\) | \(2 \text{ J}\) |
| Final | \(4 \text{ m/s}\) | \(1 \text{ kg}\) | \(8 \text{ J}\) |
Revision Table: Key Concepts for Work and Energy
| Concept | Definition/Formula | Units |
|---|---|---|
| Work Done (\(W\)) | Energy transferred by a force acting over a distance; Change in kinetic energy | Joules (J) |
| Kinetic Energy (\(KE\)) | Energy of motion | Joules (J) |
| Work-Energy Theorem | \(W_{net} = \Delta KE\) | N/A |
| Mass (\(m\)) | Measure of inertia | Kilograms (kg) |
| Speed (\(v\)) | Magnitude of velocity | Meters per second (m/s) |
Additional Information on Work, Energy, and Speed
Work done is a scalar quantity, meaning it only has magnitude, not direction. When positive work is done on an object, its kinetic energy increases (it speeds up). When negative work is done, its kinetic energy decreases (it slows down).
Kinetic energy is directly proportional to the mass of the object and the square of its speed. This means that doubling the speed of an object increases its kinetic energy by a factor of four.
The units used in these calculations are standard SI units:
- Mass is in kilograms (kg).
- Speed is in meters per second (m/s).
- Work and energy are in Joules (J). One Joule is equivalent to one Newton-meter (N·m) or one kg·m<sup>2</sup>/s<sup>2</sup>.
This problem highlights how work is the mechanism by which energy is transferred to an object to change its state of motion.
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