EK 1001 Physics Questions

[Ehwic]

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226: A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is .1, how far does the car slide? how much time is needed for the car to come to a stop?

I was wondering, are these two problems solvable with the amount of information they give you? If so, can someone explain their reasoning behind how they got that answer?

Choices for first part of the question:
a) 50m
b) 100m
c) 200m
d) 400m

Second part:
a) 1s
b) 10s
c) 20s
d) 40s

Another question: An object is placed on a spring. The spring is compressed downward and released. From the moment the spring is released until the object leaves the spring, the magnitude of the objects acceleration will:

Increase then decrease
decrease then increase

I chose the first one because the way i thought of it was that there is no acceleration when the spring is compressed. As soon as you release it from the compressed position, it would need to accelerate and then it would eventually decelerate as the distance from the relaxed position gets smaller. However, this was wrong. It actually decreases then increases. Can someone elaborate as to why this happens?

 

[pm1]

226: A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is .1, how far does the car slide? how much time is needed for the car to come to a stop?

I was wondering, are these two problems solvable with the amount of information they give you? If so, can someone explain their reasoning behind how they got that answer?

Choices for first part of the question:
a) 50m
b) 100m
c) 200m
d) 400m

Second part:
a) 1s
b) 10s
c) 20s
d) 40s

The way I solved was by figuring out the acceleration first. Since friction is what is going to cause the car to decelerate I must find what is the - acceleration from friction.
I know that Ffriction = N u (coefficient of friction).
Since that is the only force acting on the car Fnet = Ffriction

ma = N u
ma = mg u (cancel m out)
a = g u
a = 10 (0.1) = 1

Then, solve for x from V^2 = Vo^2 + 2ax
0 = (20)^2 + 2(1)x
-400/2 = x
x = 200

Then for time
V = Vo + at
0 = 20 + 1t
t = 20

 

[Ehwic]

wow thanks for the help. For some reaosn when I posted this question last night, I had no clue what the book was talking about at all. I guess I had a giant brain fart. but thanks for the help!