My math says this is wrong, but what part of energy conservation am I violating?

Question

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

Answer 1

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.

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

Answer 2

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.

Answer 3

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 $h$.

Answer 4

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!

Answer 5

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

work done by friction is force*distance or u * normal reaction* x

u is coefficient of friction and normal reaction here is mg

and x is the distance travelled in that friction enabled surface.

so using conservation of energy

work done by friction = final energy - initial energy

Answer 6

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..

Answer 7

@Vineet

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.