How much work is needed to change the speed of a 2375 kg sport utility vehicle from 15.0 m/s to 60 m/s? a. 2.2 MJ O b. 2.7 MJ O c. 2.8 MJ d. 3.1 MJ e. 4.0 MJ
How much work is needed to change the speed of a 2375 kg sport utility vehicle from 15.0 m/s to 60 m/s? a. 2.2 MJ O b. 2.7 MJ O c. 2.8 MJ d. 3.1 MJ e. 4.0 MJ
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ISBN:9781305952300
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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![### Work and Energy Calculation
**Problem:**
How much work is needed to change the speed of a 2375 kg sport utility vehicle from 15.0 m/s to 60 m/s?
**Options:**
- a. 2.2 MJ
- b. 2.7 MJ
- c. 2.8 MJ
- d. 3.1 MJ
- e. 4.0 MJ
**Explanation:**
To solve this problem, we need to calculate the change in kinetic energy of the vehicle because the work done on the vehicle is equal to the change in its kinetic energy. The kinetic energy (KE) of an object is given by:
\[ KE = \frac{1}{2}mv^2 \]
where:
- \( m \) is the mass of the object,
- \( v \) is its velocity.
Let's calculate the initial and final kinetic energies:
1. **Initial Kinetic Energy (KE_initial):**
\[
KE_{\text{initial}} = \frac{1}{2} \times 2375 \, \text{kg} \times (15.0 \, \text{m/s})^2
\]
2. **Final Kinetic Energy (KE_final):**
\[
KE_{\text{final}} = \frac{1}{2} \times 2375 \, \text{kg} \times (60.0 \, \text{m/s})^2
\]
3. **Change in Kinetic Energy (\(\Delta KE\)):**
\[
\Delta KE = KE_{\text{final}} - KE_{\text{initial}}
\]
Finally, the work done is equal to the change in kinetic energy. The correct answer will be one of the given options in megajoules (MJ).
### Practice Exercise:
Try to calculate the exact numerical values using the given formulae to find the correct option. This exercise helps in understanding the application of kinetic energy in real-world scenarios, such as calculating the energy required for accelerating vehicles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F765f1446-f94a-48d3-b4ac-193b47a28028%2F2d9c5bff-b7a4-4a5e-afa6-ee373482906a%2Fjea4gqa_processed.png&w=3840&q=75)
Transcribed Image Text:### Work and Energy Calculation
**Problem:**
How much work is needed to change the speed of a 2375 kg sport utility vehicle from 15.0 m/s to 60 m/s?
**Options:**
- a. 2.2 MJ
- b. 2.7 MJ
- c. 2.8 MJ
- d. 3.1 MJ
- e. 4.0 MJ
**Explanation:**
To solve this problem, we need to calculate the change in kinetic energy of the vehicle because the work done on the vehicle is equal to the change in its kinetic energy. The kinetic energy (KE) of an object is given by:
\[ KE = \frac{1}{2}mv^2 \]
where:
- \( m \) is the mass of the object,
- \( v \) is its velocity.
Let's calculate the initial and final kinetic energies:
1. **Initial Kinetic Energy (KE_initial):**
\[
KE_{\text{initial}} = \frac{1}{2} \times 2375 \, \text{kg} \times (15.0 \, \text{m/s})^2
\]
2. **Final Kinetic Energy (KE_final):**
\[
KE_{\text{final}} = \frac{1}{2} \times 2375 \, \text{kg} \times (60.0 \, \text{m/s})^2
\]
3. **Change in Kinetic Energy (\(\Delta KE\)):**
\[
\Delta KE = KE_{\text{final}} - KE_{\text{initial}}
\]
Finally, the work done is equal to the change in kinetic energy. The correct answer will be one of the given options in megajoules (MJ).
### Practice Exercise:
Try to calculate the exact numerical values using the given formulae to find the correct option. This exercise helps in understanding the application of kinetic energy in real-world scenarios, such as calculating the energy required for accelerating vehicles.
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