# How do you find object's velocity and height using energy?

1. Dec 7, 2016

### Sciencelover91

1. The problem statement, all variables and given/known data
In the diagram above, a box of mass .385 kg is sliding at the right at 1.55 m/s across a frictionless surface when it falls into the hole shown.
a) How fast is it moving when it passes point P?
b) How fast is the box moving when it passes through the bottom of the hole the first time? What about the second time?
c) To what vertical height does it rise on the right side?
d) How far along the 45 degree incline does it slide?
2. Relevant equations
Ug (Gravitational Potential energy) = mgh
Kinetic Energy (KE) = .5 x m x vfinal^2
vfinal^2= vinitial^2 +2a x displacement

3. The attempt at a solution
a) I did (.385kg)(9.81m/s^2)(1.25m) and got 4.72 J and set that equal to the kinetic energy equation (4.72 J = .5(.385kg)v^2 ) and got 4.96m/s^2.
However, I was told that the height I might be the displacement between the heights which is .8 to get a velocity of 3.96, I wondering why potential energy and kinetic energy are equal and why I used .8?
b) Using the 3.96 m/s from part a, I did v^2 = (3.96m/s)^2 +2(9.81m/s^2)(1.25m) = 6.34 m/s. I don't know why I use 1.25 m though and what the question means by a second time. Does it mean it goes to the right side and slide back down?
c & d) I don't know how to approach c or d so I wanted to ask for any hints please?

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2. Dec 7, 2016

### BvU

Hi lover,

I read a) is about point P, so why the 1.25 m ?
Your third relevant equation is not relevant here, but you need something about energy conservation: the initial speed also influences the speed at point P

3. Dec 7, 2016

### haruspex

... and where exactly is P? Go check.

4. Dec 8, 2016

### TJGilb

Think about conservation of energy and the work-energy theorem, in this case: $\frac 1 2 mv_i^2+mgh_i+W=\frac 1 2 mv_f^2+mgh_f$

This one looks like it should be solvable with just that equation.