- #1
OlicityFangirl
Homework Statement
Hey all, another problem. There is a 1 meter wooden panel that is on a frictionless surface. A puck is placed on one end of the panel and is pushed towards the other side. The mass of the panel is 10 times that of the puck. Also, there is a friction coefficient of uk=0.15 between the puck and the board. I have to find the minimal initial velocity that the puck must have to be able to slide off the board.
Homework Equations
Conservation of Energy: K1 + U1 + Wother = K2 + U2
Work: W=Fs
Friction Force: Fk= ukN
Kinetic Energy: 0.5mv2
Newton's Second Law: F=ma
Kinematics: x= xo + vot + 0.5at2
The Attempt at a Solution
First thing I've done is set up the conservation of energy equation. The puck's initial kinetic energy is converted to kinetic energy of the wooden panel and some is "lost" as friction (Wother). Using this logic, and adjusting the mass of the wooden panel in terms of the puck, my conservation equation is the following:
0.5mvo2 = 5mvp2 + ukmgs
I need to find the unknown vp (the velocity of the panel when the puck has crossed the entire panel). This is where I'm not sure what to do. I decided to go with kinematics, and where I end up is a huge mess and I'm not sure I can isolate vo. Here is my work/methodology:
The movement of the panel is caused by the reactionary force of the kinetic friction acting on the puck. Thus the force is the same as the kinetic friction force (mass of panel is adjusted for):
Fk,r = ukmg=10map
ap = [ukmg] / 10m = 0.1ukg
This is the acceleration, and I need to find the velocity when the puck is at x=1. Acceleration is a=vt, so I need to find how long it takes for the puck to travel 1 m. I used this as my equation for the motion of the puck on the wooden panel:
x = vot - 0.5ukgt2 = 1m
vot - 0.5ukgt2 - 1 = 0
Using the quadratic formula, I found the following solution:
t= [-vo + (vo2 - 4ukg)0.5] / ukg
I plugged this into the equation for the velocity of the panel (vp = at), getting (the 0.1 is from accounting for the mass of the panel vs puck):
vp = 0.1(-vo + (vo2 - 4ukg)0.5)
I've plugged this back into my energy conservation equation, and now the equation is a sizable mess and I'm not sure I can isolate vo. Is there any mistake in my math/logic, an easier way to find vb I haven't thought of, or am I just going to have to work through the complicated equation I'm at now?