We appreciate your visit to During an emergency stop a 1600 kg car lost tex 3 2 times 10 5 tex Joules of kinetic energy If the car traveled 25. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
The friction force acting on the car was approximately [tex]\(1.28 \times 10^4\)[/tex] N.
1. First, let's use the work-energy principle to find the friction force F:
[tex]\[ \text{Work done by friction force} = \text{Change in kinetic energy} \][/tex]
[tex]\[ F \times d = \Delta KE \][/tex]
Where:
- F is the friction force (in Newtons),
- d is the distance traveled (in meters),
- [tex]\( \Delta KE \)[/tex] is the change in kinetic energy (in Joules).
2. Given that the car lost [tex]\(3.2 \times 10^5\)[/tex] Joules of kinetic energy and traveled 25 meters:
[tex]\[ F \times 25 = 3.2 \times 10^5 \][/tex]
3. Now, let's solve for \(F\):
[tex]\[ F = \frac{3.2 \times 10^5}{25} \][/tex]
[tex]\[ F = 1.28 \times 10^4 \][/tex]
Therefore, the friction force acting on the car was approximately [tex]\(1.28 \times 10^4\)[/tex] N. Hence, the correct option is B) [tex]\(1.28 \times 10^4\)[/tex] N.
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