### Physics Problem### Problem Statement:1) How much work is needed to change the speed of a 1837 kg sport utility vehicle from 15.0 m/s to 60 m/s? (1 MJ = \(10^6\) Joules)### Multiple Choice Options:a. 1.10 MJ b. 2.7 MJ c. 2.2 MJ d. 3.1 MJ e. 4.0 MJ *Note:* There are no graphs or diagrams included in this problem statement. It consists solely of the question and multiple-choice answers.### Explanation:This problem involves calculating the kinetic energy change required to accelerate the vehicle and thus the work done on the vehicle. The formula for kinetic energy (K.E) is given by:\[ K.E = \frac{1}{2}mv^2 \]Where:- \( m \) is the mass of the vehicle (1837 kg)- \( v \) is the velocityThe work done (\( W \)) to change the speed of the vehicle can be determined by finding the difference in kinetic energy at the two speeds:\[ W = \Delta K.E = K.E_{\text{final}} - K.E_{\text{initial}} \]\[ K.E_{\text{initial}} = \frac{1}{2} \times 1837 \times (15)^2 \]\[ K.E_{\text{final}} = \frac{1}{2} \times 1837 \times (60)^2 \]Substitute the values to find the kinetic energy at both speeds, then find the difference to determine the work done.

Given data: Mass (m) = 1837 kg Initial velocity (Vi) = 15.0 m/s Final velocity (Vf) = 60 m/s…

Answered: 1) How much work is needed to change…

1) How much work is needed to change the speed of a 1837 kg sport utility vehicle from 15.0 m/s to 60 m/s? (1 MJ = 106 Joules) a. 1.10 MJ Ob.2.7 MJ O C. 2.2 MJ O d. 3.1 MJ e. 4.0 MJ

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Instantaneous Power

Power is defined in physics as the amount of energy transmitted or transformed per unit time. The watt is the unit of electricity in the International System of Units, equivalent to one joule per second.

Question

### Physics Problem

### Problem Statement:
1) How much work is needed to change the speed of a 1837 kg sport utility vehicle from 15.0 m/s to 60 m/s? (1 MJ = \(10^6\) Joules)

### Multiple Choice Options:
a. 1.10 MJ
b. 2.7 MJ
c. 2.2 MJ
d. 3.1 MJ
e. 4.0 MJ

*Note:* There are no graphs or diagrams included in this problem statement. It consists solely of the question and multiple-choice answers.

### Explanation:

This problem involves calculating the kinetic energy change required to accelerate the vehicle and thus the work done on the vehicle.

The formula for kinetic energy (K.E) is given by:
\[ K.E = \frac{1}{2}mv^2 \]

Where:
- \( m \) is the mass of the vehicle (1837 kg)
- \( v \) is the velocity

The work done (\( W \)) to change the speed of the vehicle can be determined by finding the difference in kinetic energy at the two speeds:

\[ W = \Delta K.E = K.E_{\text{final}} - K.E_{\text{initial}} \]

\[ K.E_{\text{initial}} = \frac{1}{2} \times 1837 \times (15)^2 \]
\[ K.E_{\text{final}} = \frac{1}{2} \times 1837 \times (60)^2 \]

Substitute the values to find the kinetic energy at both speeds, then find the difference to determine the work done.

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