 #1
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Homework Statement
Two pucks (5 kg each) made of Teflon are on a long table, also made of Teflon. Puck A is sitting at
rest on the left end of the table. Puck B is 15 m away at the right hand end of the table, and is
travelling toward Puck A with an initial speed of 0.5 m/s. A person on the left waits 1.5 seconds and
then pushes Puck A forward from rest with a force of 20 N applied at an angle of 30° to the horizontal.
If that force is maintained at a steady rate for the entire question, how far will puck A travel before it
strikes puck B? (The coefficient of friction between the Teflon and Teflon is 0.04.
2. The attempt at a solution:
Puck A
Fax = 20NSin30
Fax = 10N
Fay = 20NCos30
Fay = 17.32N
Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N
Fn = Fay + Fg
Fn =17.32N + 49N
Fn = 66.32
Ff = uFn
Ff = (0.04)(66.32N)
Ff = 2.65
Fnet = Fax  Ff
Fnet = 10N  2.65
Fnet = 7.35N
a = Fnet/m
a = 7.35N/5kg
a = 1.47m/s^2
d = v1t + 1/2at^2
15x = 1/2(1.47)(y1.5)^24
x = 0.74y^2 + 2.22y +13.33
Puck B
Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N
Ff = uFn
Ff = (0.04)(49N)
Ff = 1.96N
Since there is no applied force Fnet = Ff
a = Fnet/m
a = 1.96/5kg
a = 0.39m/s^2
d = v1t + 1/2at^2
x = (0.5)(y) + 1/2(0.39)(y)^2
0.74y^2 + 2.22y +13.33 = 0.5  0.195y^2
0 = 0.545y^2  1.72y13.33
I then put it in the quadratic formula and got:
y = 6.77s and y = 3.61s(omited)
Puck A Distance
d = v1t + 1/2at^2
d = (0m/s)(6.77s1.5s) + 1/2(1.47m/s^2)(6.771.5s)^2
d = 20.41m(This answer doesn't make sense)
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