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faust18 [17]
3 years ago
8

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

ce against the wheels?
Physics
1 answer:
Artemon [7]3 years ago
6 0
Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks. 

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)
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Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

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