Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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**Physics Problem: Kinetic Energy and Force Calculation**

A driver initially at rest accelerates her 2000 kg car to a final velocity of 30 m/s.

a. **What is the kinetic energy of the car at her final velocity?**

To find the kinetic energy (\(KE\)) of the car at its final velocity, use the formula:

\[ KE = \frac{1}{2} m v^2 \]

Where:
- \( m \) is the mass of the car (2000 kg)
- \( v \) is the final velocity (30 m/s)

b. **The car travels for a distance of 275 meters. How much force is being applied to the car?**

To find the force (\(F\)) applied to the car, use the formula derived from work-energy principle:

\[ W = F \cdot d = \Delta KE \]

Where:
- \( W \) is the work done,
- \( d \) is the distance (275 meters),
- \(\Delta KE\) is the change in kinetic energy (since the car starts from rest, \(\Delta KE\) is the final kinetic energy calculated from part a).

The solution involves computing the work done and equating it to force times distance to solve for the applied force.
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Transcribed Image Text:**Physics Problem: Kinetic Energy and Force Calculation** A driver initially at rest accelerates her 2000 kg car to a final velocity of 30 m/s. a. **What is the kinetic energy of the car at her final velocity?** To find the kinetic energy (\(KE\)) of the car at its final velocity, use the formula: \[ KE = \frac{1}{2} m v^2 \] Where: - \( m \) is the mass of the car (2000 kg) - \( v \) is the final velocity (30 m/s) b. **The car travels for a distance of 275 meters. How much force is being applied to the car?** To find the force (\(F\)) applied to the car, use the formula derived from work-energy principle: \[ W = F \cdot d = \Delta KE \] Where: - \( W \) is the work done, - \( d \) is the distance (275 meters), - \(\Delta KE\) is the change in kinetic energy (since the car starts from rest, \(\Delta KE\) is the final kinetic energy calculated from part a). The solution involves computing the work done and equating it to force times distance to solve for the applied force.
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Can you elaborate on answer ? Also what would be the answer for question b? I don't understand your explanation. Thank you. 

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