A student hits a hockey puck, giving it an initial velocity of 6.0 m/s. If the coefficient of kinetic friction between ice and puck is 0.100, how far will the puck slide before stopping? This is impossible because it only gives you merely two pieces of information! I swear to god, this is the sole reason why physics is terrible! This subject is just useless and it sucks!
+1224 mass = 0.25 kg
initial velocity = v_i = 6 m/s
coefficient of kinetic friction = mu_k = 0.1
final velocity = v_f = 0 m/s
distance = d = ?
There are three forces acting on the puck (friction, gravity, normal force) Gravity and normal force cancel each other out, so we can ignore that for now.
Friction acts in the -x direction.
F_n = mg = (0.25 kg)(9.8 m/s^2) = 2.45 N
F_f = (mu_k) * (F_n) = (0.1) * (2.45) = 0.245 N
F = ma ==> a = F/ m = (0.245 N) / (0.25 kg) = 0.98 m/s^2
Finally, we use kinematics.
(v_f)^2 = (v_i)^2 + 2ad
0 = (6^2) + 2(0.98)d
d = 18.367 m
My physics is a bit rusty so this is the best I can come up with.
+6 If the question only gives you two pieces of information, you can try using variables to sub for the other variables you'll need to solve for the question. Most likely, the variables will cancel out, leaving you with the answer.
people are saying the μmg = ma. Why the h**l is the ma there?? The force of friction doesn't fucking equal the net force!!
+6 I haven't gotten up to that part in my physics classes yet, but you should try using variables to sub for the values you don't know.
Right now, what you have is this:
initial velocity: 6.0 m/s
final velocity: 0 m/s
Coefficiant of kinetic friction: 0.100
Draw a force diagram. Since it looks like you have more information on velocities, you might be able to find acceleration or something like that, use a few variables and more equations, and solve for the answer.
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+36916 Force of friction = coefficient of friction * NORMAL force Normal force = m a = mass * 9.81 m/s^2
were you not given the mass of the hockey puck?
+2440 With the given parameters, the mass of the hockey puck is NOT required to solve this problem.
However, intelligence slightly above that of a hockey puck IS required to solve this problem.
It’s great to have you back, EP. You are my favorite member to troll. 
GA
+36916 Thanx Troller ! Yah....I did not look very far in to this problem....I suspected 'm' may cancel out in there, but my pucked brain doesn't have time right now...... water pipes are leaking ......grrrrr!
~EP
+36916 x = x0 + v0t + 1/2 at^2
f friction = u m g (using g to avoid confusion)
ff = ma and = umg
a = u g
vf = vo + at
0 = 6 + u g t
0 = 6 - .1 (9.81) t t = 6.12 sec
x = x0 + vo t - 1/2 at^2
= 0 + 6(6.12) - 1/2 (.1)(9.81)(6.12)^2 = 18.35 meters
NOW ---- off to fix those leaks.......
