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Basically energy of conservation says it doesn't matter what path I take, energy will be the same from before as it is after.

So I took the block at the top's energy equal to the energy when it lands on the ground (when it stays still)

$$E_i = mgh - \mu mg(1.3)$$ and $$E_f = 0$$, Energy is 0 because it s on ground level and not moving

$mgh - \mu mg(1.3) = 0$

But I get something absurd like $h = 0.39m$

Which makes no sense at all

Now if I had set it equal to the point where it doesn't fly off I get

$mgh - \mu mg(1.3) = mg(1.9)$

Then I get h = 2.29m. Which is A LOT better.

So what part of energy of conservation did I violate in the beginning? This isn't homework, this is just self-study

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energy is not zero at ground, because you have to take the evergy just before it hits the ground. not after it had hit. so your first equation will become, $mgh - \mu mg(1.3) E_f = \frac{1}{2}mv^2$ – Vineet Menon Dec 13 '11 at 5:23
What do you mean? It will stop eventually, can't I use "that state when it stops" into my equation? – jak Dec 13 '11 at 5:39
because a lot of energy is lost when it stops, the ground has to absorb all the kinetic energy to stop the block, and thus your "box-slope system" is no longer isolated. – Vineet Menon Dec 13 '11 at 5:40
2.29m isn't even close. 5.002m is the correct answer. – Casey Robinson Dec 13 '11 at 18:19
@Casey: although there's nothing wrong with pointing out when the answer is wrong, just remember that this question is more about the concept than getting the right answer. – David Z Dec 13 '11 at 19:21

I wanted to write it as a comment but for some reason 'add comment' box isn't there.

In my opinion the mistake you did is that you ignored the 'Kinetic Energy', the block must have some speed if it slips down from a height.

$$K_{1} + U_{1} +W_{other} = K_{2} + U_{2}$$

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But the block starts from rest – jak Dec 13 '11 at 5:13
But It does have some Velocity when it reaches the end, What I am trying to tell you is that the block has some final Kinetic Energy. Imagine a ball dropped from a height h above the ground, It's initial velocity is Zero but the final velocity is not Zero! – Ishaan Singh Dec 13 '11 at 5:39
@IshaanSingh: U don't have enough privilege to add comment since your repo is low...once you gain enough repo, you can comment for any answer...Till then u have to suffer... – Vineet Menon Dec 13 '11 at 8:38
@CaseyRobinson I was just pointing out his mistake, He already accounted for 'friction' in both of his equations but forgot to consider 'Kinetic Energy' of the block. Thanks for bringing this out by the way, It would avoid confusion for other users. – Ishaan Singh Dec 14 '11 at 2:12
@IshaanSingh Thanks for the explaination. I gave you a vote and deleted my comment since it was solved. – Casey Robinson Dec 14 '11 at 11:14

Your approach is incorrect. The key information you are given is the horizontal point of impact, from which you can calculate the horizontal velocity of the block as it leaves the ramp. Then set the equivalent kinetic energy equal to the energy gained descending the slope, minus the energy lost to friction and solve for the initial height.

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First you know that when the block reaches the end of the track it doesn't have vertical velocity and it falls freely under the gravity so you can calculate the time as $1.9m=\frac{1}{2}\times 9.81m/s^2 t^2$ so you know the time of flight. Then using the horizontal length you could calculate the horizontal component of the speed by $4m=v_h \times t$ Then you know the horizontal component of the speed $v_h$ and the vertical is $v_t=9.81 \times t$ and you can calculate the total speed when it hits the ground by $v=\sqrt{v_t^2+v_h^2}$

To calculate the $h$ just use the conservation of energy

  1. PE at height $h$ is $mgh$
  2. KE before sliding over the green surface is $mg(h-1.9)$
  3. You can calculate the KE and PE just before the block leaving the track, $PE=528g \times 9.81m/s^2 \times 1.9m$ and $KE=0.5 \times m \times v^2$
  4. You need to calculate the energy loss on the green surface due to friction.

Write down the equation for energy conservation and solve for $h$.

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Why does it have a vertical velocity? I thought it undergoes free fall – jak Dec 13 '11 at 18:34
@CaseyRobinson: Your definition of free fall is not the standard one. Free fall is, strictly speaking, used for objects that are accelerating under the influence of gravity in the absence of other forces. Your definition is closer to that of terminal velocity. – qubyte Dec 15 '11 at 16:51

I think you just neglected the kinetic energy which was gained by the body coming from 'h' height to a height '1.9m' measured from the ground.Check it!

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The friction also needs to be accounted for. – Casey Robinson Dec 13 '11 at 18:25

You did not consider the work done by friction in the equation for energy.

$$W_F =F_Fx = uF_Nx$$

$u$ is coefficient of kinetic friction and $F_N$ is the normal force $mg$ and $x$ is the distance travelled on the surface.

Conservation of energy states that:

$W_F=\delta E=E_f-E_i$

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So, this is an attempt to give you a consolidated answer,

In classical mechanics problem like this it is conventional to use either of the two conservation laws viz z viz Energy and Momentum. Conservation of energy principle is applicable only in a closed system, where there is no escape of matter nor energy. You can consider a thermos flask( ignoring the minute heat escape through radiation) as a closed system. Whereas momentum conservation necessitates only matter shouldn't escape the confinement of the system. I'm talking in a purely classical terms. No quantum theory here.

Now, coming back to your question. In your question, you want to apply conservation of energy. For that you should first define your system, which is closed for energy as well as matter. So, I would take the slope-box as my system.

From here I can enumerate the energy gains and losses at different stages of the trajectory of box...

Initial : Full potential Energy...

Magnitude: $mgh$

Final before Collision: Full kinetic Energy - some loss due to friction

magnitude: $\frac{1}{2}mv^2$

Final after collision: Zero Energy.

magnitude: $0$, but energy has been transferred to earth, dissipated as sound, light(probably) and heat.

Since after collision, there is many unaccountable energy, we take the initial energy and final energy before collision. (PS. there's no harm in taking energy after collision, but then how to measure the various energies?? Anyway, we are interested in the final velocity).

$$Initial = Final$$ $$mgh - \mu mg (1.3)= \frac{1}{2}mv^2$$

This equation requires two independent variables. So, this approach won't work. You have to find out the time of flight of the box and then from that get the velocity from the standard kinematic equations..

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have u tried @user6629's approach? – Vineet Menon Dec 15 '11 at 5:12


I did it first with kinematics to the speed when it leaves the track. Even with velocity, I didn't get 5.0002m as the height.

$t = \sqrt(1.9/4.9) = 0.623s$

$4/0.6235s = 6.42m/s$

$mgh - \mu mg (1.3)= \frac{1}{2}mv^2 + mg(1.9)$

Solving, I still get h = 4.39m.

Also, "total speed" isn't a proper physics term. It should (for part c) ask "the speed" right? or "total velocity"? Because that's the magnitude.

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