Conservation of Energy with Friction: Where Does the Block Stop?

[natedigity]

[SOLVED] conservation of E w/ friction

a block is released from rest at height d = 58 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.54. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.54. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop? If its final stop is on a plateau, state which one and give the distance L from the left edge of that plateau. If the block reaches the ramp, give the height H above the lower plateau where it momentarily stops.

(a)The block stops on the: 1 - First plateau; 2 - Second plateau; 3 - Ramp. Give the number of the correct answer.

(b) Give L or H, which ever is appropriate.


Part a) I know it will not stop on the ramp and if it was frictionless on the horizontal surfaces it would end up on the lower plateau. I don't know how to figure the kinetic friction into the equation. The mech energy is constant, so ke and pe will change based on the height and angle and the kf will factor in there somewhere.


Initial PE is mgd
I have no mass, or angle?
deltaEth = Fkd = PEmgd
Emech2=Emech1-Fkd
KE = deltaPE + PEmgd = mg(d+PEd)?
I am not understanding how to set this up to find where it stops except it is where all energy has been lost to friction.

 

[HallsofIvy]

Several things are missing here: First, you say "a block is released from rest at height d = 58 cm". Height above what? The first plateau or the second? Also, you do not give the mass of the block. If there were no friction, it wouldn't matter since the mass would cancel out of everything

You can't use "conservation of energy" on the plateaus. With friction, mechanical energy is not conserved (you would have to include the heat produced by the friction and I don't see any good way to do that). You can use conservation of energy to find the speed of the block at the bottom of each plateau and then use F= ma (if the mass is m, its weight is mg so the friction force is -mg(.54). a= .54g.) to calculate how much the block slows while crossing a plateau.

 

[natedigity]

d is the height of the entire ramp, as well as the distance across the first plateau. There is no mass, or speed, or angle given. I understand the energy is not conserved, it is lost to friction. That is what is confusing me on finding the distance it will travel across the friction surface.

 

[Shooting Star]

More infmn is required. If the vertical dist from the starting point to the first plateau is h, and h< 0.54*d, then the block will stop on the first plateau. Until that is known, you can't proceed further.

Is the 2nd plateau at ground level? Then it can be solved.

 

[Shooting Star]

The mass doesn't matter. If d is the height of the initial pt above the 2nd plateau, then it can be solved. Angles, speed is not required if it is starting from rest.

 

[natedigity]

d is the height to the top, d/2 is the distance to the first plateau, and the second plateau is on ground level.

 

[natedigity]

also the length of the first plateau is d as well. I think it will stop on the first plateau, but part b) asks how far it will travel on the plateau, would that be .54*d because of the friction?

 

[Shooting Star]

From the new infmn you have given, the vertical dist from top to the 1st plateau is d/2. I’ll denote the co-eff of friction by k. Then frictional forces acting on the plateaus will be kmg in the direction opp to motion.

Suppose v1 is the speed at the end of the first frictionless ramp, that is, when it is about to go on the 1st plateau. Equate the PE at the top to the KE at this point.

Then, mv1^2/2 = mg(difference of height) = mg(d/2) ==> v1^2 = 2gd/2 = gd.

Now it is traveling on 1st plateau with friction. Let ‘a’ be the deceleration.. Let the speed at the end of the plateau be v2, and initial speed is v1. Also, ma= kmg ==> a = kg.

Now use the formula for deceleration for a dist of d with deceleration = -a.

Then, v2^2 = v1^2 -2ad = gd -2kgd = gd( 1 - 2k). = gd (1 – 2*0.54) < 0.

Since v2^2 cannot be negative, this means that the block has not reached the end of 1st plateau. So it stops somewhere on the first plateau.

I’m sure you can take it up from here. Try before asking again.

 

[natedigity]

thank you, that makes sense now.