Deriving Formulas for Work/Energy Problems

[CL1349]

Homework Statement

An ice cube is given a push at the top of an incline giving it an initial speed of 7 m/s down the incline. At the bottom of the incline the ice cube is deflected vertically upward as shown in the accompanying diagram. Ignoring friction and air resistance, how high H does the ice cube go above the floor before falling back down?

Note: Solve using work/energy considerations

The problem with a graph of it is attached.

Homework Equations

W = Ef - Ei

Ef = Ei

The Attempt at a Solution

I set the work equal to zero, and set the initial energy equal to the final energy. I worked the equation around and came up with

mgh = 1/2 mv^2 + mgh

mg(4m) = 1/2 m (Vf) ^2 + mg( 0 m)

I solved for final velocity with Vf = square root( 2gh), square root (2 * 9.8 m/s2 * 4 m)

I got 8.85 m/s for final velocity. I don't know if this is right or not. But how can I rearrange the work/energy theorem to find the H?

 

[obafgkmrns]

You're making it too difficult. The sum of the initial potential energy (mgh) and the initial kinetic energy (mv^2) is equal to the final potential energy (mgH) when the ice cube reaches its highest point. (Keep in mind that the ice cube's speed and kinetic energy are zero at that point.) You never have to worry about the ice cube's speed after it's released.

 

[CL1349]

Thanks for the help obafgkmrns. You really made me see how I can easily get confused on such a simple thing.

It's hard for me to figure out how to analyze problems like this and derive the right formulas cause I'm not that great at physics. And it doesn't help that my professor is a bit looney.

So, here's my attempt at the solution.

Initial PE + KE = Final PE + KE

the initial KE = 1/2 mv^2.

1/2 * m * (7 m/s) ^2

= 24.5 m

The initial PE = mgh

m * 9.8 * 4 m

= 39.2 m

63.7 m = mgh + 1/2 mv^2 , v = 0


h = 63.7m/mg

h = 63.7/9.8

h = 6.5 meters