Solving Hockey Puck Problem Homework

[chiurox]

Homework Statement

A hockey puck with a mass of 0.50 kg is hit in a straight line along a horizontal playing surface.
Well, there are 4 parts to this question:
a. If the puck has an initial velocity of 4.0m/s and stops after 5.0, what is the coefficient of friction, uk1, of the playing surface?
My answer: for this one, I got 0.16 and I'm quite sure it's right, unless someone can prove me wrong.

b. What should the initial velocity be if the coefficient of friction is 0.16 and the puck has to travel 6.0m?
My answer: I got 4.4m/s and again I think it's right, unless someone can prove me wrong.

c. Someone spills food on the playing surface, changing the coefficient of friction over a 1.0 m section. Now, in order for the puck to travel 6.0 m, it first goes 4.0 m over a surface with a coefficient of friction of 0.16, then it goes 1.0m over a surface with a coefficient of friction uk2, and finally it goes 1.0 m over a surface with a coefficient of friction of 0.16 again. If the initial velocity of the puck in a straight line must now be 5.0 m/s in order to travel 5.0m, what is the coefficient of friction, uk2?
My answer: I'm stuck on this one... help please?
d. What must the initial velocity of the puck in (c) be in order for the puck to stop after 6.0m?
Well, since I'm having problems with (c), I don't know about this one.


Could anyone help me please as soon as possible?
Thanks

 

[learningphysics]

Parts a and b look right to me.

have you studied work/energy yet? the next parts can be done with work/energy... it is much easier than the kinematic way described below.

For part c).

you have 2 parts:

v0=5m/s<----4.0m---->v1<----1.0m---->v =0

so over 5.0 m... the first 4.0m is a surface of coeff. 0.16. the next 1.0m is uk2.

find the velocity v1 just as it gets on the part of the surface with uk2 (you can get this using the initial velocity v0, distance=4.0m and acceleration, which you can get from the coefficient of friction 0.16).

now you can get the acceleration over the uk2 surface using d = 1.0m final velocity v= 0... and v1 which you just calculated above... from that you can find frictional force and uk2.

for part d). work backwards...
you have 3 parts

v0 <---- 4.0m ----> v1 <---1.0m----> v2 <---1.0m---> v = 0

get v2... then v1... then v0...

 

[chiurox]

I'm not sure if I got what you told me, but I worked out the problem in a similar way:
Vo^2 / 2(uk2)(g) = 5.0 which is the distance traveled,so I got the coefficient of uk2 = 0.25
Well, since the puck has to go over a total distance of 5 meters with a velocity of 5.0m/s, and we know from part (a) that with an initial velocity of 4.0m/s the puck stopped after 5.0m. So uk2 has to be significantly bigger than uk1 since the puck is traveling the same distance (5m) but with a velocity of 5.0m/s.

(d). I used 0.25=uk2 to plug in the equation Vo^2 / 2uk2(g) = distance traveled or 6
then I found Vo = 5.5m/s so that the puck will stop after 6.0m
I don't know if it's right, but I feel that it at least makes sense because the puck has to travel 1 more meter and therefore the initial velocity has to be a bit bigger.

Could anyone help me check or point out some mistakes I did?
I'm taking AP Physics B by self-studying, with no textbook, and doing activities online (such as this one) worth grade. So I'm pretty much lost almost all of the time heh...