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

• Sciencelover91
In summary: So plugging in the values, you get:a) The speed is 4.72 J when it passes point P.b) It is moving at 4.96 m/s when it passes through the bottom of the hole the first time.c) It rises to a height of 3.96 m on the right side (from part a).d) It slides for a total of .8 m down the incline before it falls into the hole again.
Sciencelover91

## Homework Statement

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?

## Homework Equations

Ug (Gravitational Potential energy) = mgh
Kinetic Energy (KE) = .5 x m x vfinal^2
vfinal^2= vinitial^2 +2a x displacement

## 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?
[/B]

#### Attachments

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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

Sciencelover91 said:
1.25m
Sciencelover91 said:
a) How fast is it moving when it passes point P?
... and where exactly is P? Go check.

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.

## 1. How do you calculate an object's velocity using energy?

To calculate an object's velocity using energy, you must first know the object's mass and the potential and kinetic energy at a certain point. Then, you can use the equation KE = 1/2 * m * v2 to solve for the velocity (v).

## 2. Can you find an object's velocity and height using only energy?

Yes, it is possible to find an object's velocity and height using only energy. This can be done by setting the potential energy equal to the kinetic energy at a certain point and solving for the variables.

## 3. What is the relationship between energy and an object's velocity and height?

Energy and an object's velocity and height are directly related. As an object's velocity and height increase, its kinetic and potential energy also increase. Similarly, as an object's velocity and height decrease, its kinetic and potential energy decrease.

## 4. Can you find an object's velocity and height at any point using energy?

Yes, using energy, you can find an object's velocity and height at any point. However, this method assumes that there is no external force acting on the object and that all energy is conserved.

## 5. Is finding an object's velocity and height using energy more accurate than using other methods?

It depends on the situation. In some cases, using energy to find an object's velocity and height can be more accurate, especially if there are no external forces acting on the object. However, in other cases, other methods such as using equations of motion may be more accurate.

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