Asked by Mary-Lynn
A car (1100kg) skids up a hill inclined at 6 degrees. Its initial speed is 20m/s and skids 50m before stopping.
When it stops, how high is it above where it started?
How much potential energy did it gain?
How much energy was lost to friction?
What is the force of friction (magnitude and direction)?
What is the coefficient of friction between the car and the road?
When it stops, how high is it above where it started?
How much potential energy did it gain?
How much energy was lost to friction?
What is the force of friction (magnitude and direction)?
What is the coefficient of friction between the car and the road?
Answers
Answered by
Chanz
a)50sin6 = h (trig)
b) GPE = mgh
c) KEo - GPE
KEo = 1/2 m v^2 = 1/2* 1100* 20^2
d) a = v^2/2x = 20^2/(2*50)
Ff = ma, parallel hill
e)Fn = mg cos6
mu = Ff/Fn
b) GPE = mgh
c) KEo - GPE
KEo = 1/2 m v^2 = 1/2* 1100* 20^2
d) a = v^2/2x = 20^2/(2*50)
Ff = ma, parallel hill
e)Fn = mg cos6
mu = Ff/Fn
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