A board AB of length l=1 m sits on the long, frictionless plane. A small puck is

Question

A board AB of length l=1 m sits on the long, frictionless plane. A small puck is placed at

the one end (A) of the board and pushed towards the other end (B) of the board. Mass of

the board is n=10 times the mass of the puck, coefficient of friction between the puck

and the board k=0.15. What should be the minimal initial velocity Vmin of the puck so it

will slide off the board?

http://snag.gy/SWNyf.jpg

The diagram is above, listed as Problem 1.

I have 4 other similar questions you can answer for Chegg points named "Simple Physics Question A"... "Simple Physics Question E."

Thanks for your time, and I look for complete, detailed answers with originality. :)

Explanation / Answer

The key things here are as follows.

1. The puck starts at v0 and decelerates due to the force of friction with the board.

2. The board accelerates from an initial velocity of zero due to friction with the moving puck, causing the end of the board to move forward from its original position of x=1 m.

3. We want to calculate v0 such that the puck gets to the moving far end of the board before friction slows the puck to a velocityy of zero.

Other facts: mass of board is 10mp where mp is mass of puck. Coef of friction puck to board is 0.15

The idea will be two write down the equations of motion for the puck and the far end of the board and find the shortest time when then have a common position. We can then back out the initial velocity of the puck.

equation of motion of the puck

The puck is subject to a friction force of F=0.15g*mp. SO the acceleration of the puck is

a1=0,15g. It is in the negative x direction

v1=v of puck = v0-0.15g*t

We want minimum v0 so that the puck falls off the end of the board. At that time puck will have velocity of zero so we can calculate the time this happens as

0=v0-0.15g*te or

Equation 1 : te = v0/(0.15g)

xp is position of puck. x0 = 0 (puck starts at x=0)

Equation 2: xp(t)=v0*t - (1/2)(0.15g)t^2

equation of motion for the far end of the board.

it is subgject to friction of 0.15g*mp =mb*ab where mb is mass of the board and ab is acceleration of the board, Thus

acceleration of board = 0.15g*mp/mb=0.15g*mp/(10*mp)=0.015g

So the equation of motion of far end of board is

Equation 3: xb(t)=1+(1/2)*0.015g*t^2

So now we set xp=xb and substitute in te = v0/(0.15*g) to find v0.

First we set equation 2 = equation 3

Equation 4 : xp=xb= v0t - (1/2)*(0.15g)*t^2 = 1 + (1/2)*0.015g*t^2

Substitute Equation 1 in for t in Equation 4

v0*(v0/0.15g)-(1/2)*(0.15g)*(v0/0.15g)^2 = 1 + (1/2)*0.015g*(v0/0.15g)^2

This simplifies to

v0^2/(0.3g)=1+v0^2/(20g)

Solve for v0^2

v0^2=6*g/19.7 = 6*9.8/19.7=2.98

take the square root of both sides and we get

v0=1.73 m/s which is the answer.

I hope this helps.

Remember - the more problems you do the better you will get!

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