How Does Conservation of Energy Apply to Motion Down a Hill?

[mike_302]

Homework Statement

My attempt at a solution was to say:

E_ki = E_kf + E_g

and my Ek 's all has 1/2 on them... other than that, my solution is essentially the same thing.
I end up with sqrt( v_i² - sg(delta)h ) = v_f



What's up with this?

 

[Delphi51]

The solution in the pdf file looks good and has the same answer as in your post (except for an extra "s" which must be a typo). What exactly is the question you are asking?

 

[kerimek]

I can understand your confusion with this equation. Let me break it down for you. The first part of the equation, Eki, represents the initial kinetic energy of an object. This can be calculated using the formula 1/2 * mass * velocity^2. The second part, Ekf, represents the final kinetic energy of the object at the bottom of the hill. This can also be calculated using the same formula, but with the final velocity of the object.

The third part, Eg, represents the potential energy of the object due to its position on the hill. This can be calculated using the formula mass * gravity * height. When we combine these three parts, we get the equation Eki = Ekf + Eg, which represents the conservation of energy principle. This means that the total energy of the object (kinetic energy + potential energy) remains constant throughout its motion.

Now, let's talk about the second part of your solution, where you have used the square root function. This is because the equation you have used, sqrt(vi^2 - sg(delta)h) = vf, represents the final velocity of the object at the bottom of the hill. This equation is derived from the conservation of energy principle, where we equate the initial kinetic energy to the final kinetic energy plus the potential energy gained due to the height of the hill.

In conclusion, your solution is correct and represents the conservation of energy principle in a simple and concise manner. I hope this clarifies any confusion you may have had. Keep up the good work!