Conservation of energy in x vs y direction

[Wes Turner]

Homework Statement

A child goes down a water slide. The slide is 9.0 m tall. She pushes off with an initial speed of 2.0 m/s (in the horizontal direction). If the slide is frictionless, how fast will she be going at the bottom of the slide?

Homework Equations

In the book, they use a conservation of energy equation:
K + Ug = K0 + (Ug)0

The Attempt at a Solution

(1/2)*m*v^2 + mgy = (1/2)*m*v0^2 + mgy0

Since y = 0, we have
(1/2)*m*v^2 = (1/2)*m*v0^2 + mgy0

Solving for v1, we get
v1 = sqrt(v0^2 + 2gy0)
v1 = sqrt((2.0 m/s)^2 + 2(9.8m/s^2)*(9.0m)) = 13 m/s

My problem is that the only force here is gravity and it is in the x direction, but the increase in velocity is in the orthogonal y direction. I understand that the normal force from the slide pushes the child in the x direction, but why is it not necessary to break all that down into x and y components? It seems like gravity is acting perpendicular to the vertical.

 

[DrClaude]

Energy is a scalar, not a vector, so there is no way to break it into different components. You would need to use x and y components if you were using, for instance, conservation of momentum instead.

 

[jbriggs444]

My problem is that the only force here is gravity and it is in the x direction, but the increase in velocity is in the orthogonal y direction. I understand that the normal force from the slide pushes the child in the x direction, but why is it not necessary to break all that down into x and y components? It seems like gravity is acting perpendicular to the vertical.

The portion of the slide where gravity succeeds in acting is not horizontal. It slants downward. The force of gravity and the normal force are not perpendicular.

However, as @DrClaude points out, all of that is irrelevant. A conservation of energy argument works out regardless of what angles and components are involved.

 

[Wes Turner]

The portion of the slide where gravity succeeds in acting is not horizontal. It slants downward. The force of gravity and the normal force are not perpendicular.

Ah, good point. The initial velocity was horizontal, but after that, there is both an X and Y component of the forces (not the energy).

However, as @DrClaude points out, all of that is irrelevant. A conservation of energy argument works out regardless of what angles and components are involved.

So either the object can move or it can't. If it can, then the velocities have to be whatever they are to conserve energy. Right?

Thanks

 

[jbriggs444]

So either the object can move or it can't. If it can, then the velocities have to be whatever they are to conserve energy. Right?

Right.

 

[Ray Vickson]

Homework Statement

A child goes down a water slide. The slide is 9.0 m tall. She pushes off with an initial speed of 2.0 m/s (in the horizontal direction). If the slide is frictionless, how fast will she be going at the bottom of the slide?

Homework Equations

In the book, they use a conservation of energy equation:
K + Ug = K0 + (Ug)0

The Attempt at a Solution

(1/2)*m*v^2 + mgy = (1/2)*m*v0^2 + mgy0

Since y = 0, we have
(1/2)*m*v^2 = (1/2)*m*v0^2 + mgy0

Solving for v1, we get
v1 = sqrt(v0^2 + 2gy0)
v1 = sqrt((2.0 m/s)^2 + 2(9.8m/s^2)*(9.0m)) = 13 m/s

My problem is that the only force here is gravity and it is in the x direction, but the increase in velocity is in the orthogonal y direction. I understand that the normal force from the slide pushes the child in the x direction, but why is it not necessary to break all that down into x and y components? It seems like gravity is acting perpendicular to the vertical.

If gravity acts in the x-direction, why do you write the difference in gravitational potential energy as mgy - mgy0?