Solving Pe + Ke Problem for Block of Mass m and Height h

[vysero]

A block of mass m and height h slides without friction up a hill rise of height (3/2)h as shown. In order to make it to the top of the hill, the block must have a minimum initial speed of:



Pe+Ke = Pe(f) + Ke(f)



I tried figuring out what the velocity of the block would be if it had slid down the hill. I was assuming it would require that amount of velocity to go back up the hill (maybe I assumed wrong). The answer listed is (3gh)^1/2 However my answer was (2gh)^1/2. The block has a height h but I don't understand why the size of the block should matter. Nor do I have any idea of how to incorporate that fact into my equation, please help.

 

[SammyS]

[ b]A block of mass m and height h slides without friction up a hill rise of height (3/2)h as shown. In order to make it to the top of the hill, the block must have a minimum initial speed of:[/b]

[ b]Pe+Ke = Pe(f) + Ke(f)[/b]

[ b]I tried figuring out what the velocity of the block would be if it had slid down the hill. I was assuming it would require that amount of velocity to go back up the hill (maybe I assumed wrong). The answer listed is (3gh)^1/2 However my answer was (2gh)^1/2. The block has a height h but I don't understand why the size of the block should matter. Nor do I have any idea of how to incorporate that fact into my equation, please help. [/b]

Please don't use a bold font when posting a thread.

You're right that the height of the block doesn't matter.

Check you algebra.

Show your work.

 

[vysero]

Um, when you start a thread it says (without the forward / in the beginning):

[/b]1. Homework Statement [/b]
[/b]2. Homework Equations [/b]
[/b]3. The Attempt at a Solution [/b]

So, the site is asking me to put it in bold, that's why I did it.

In any case here is my algebra in steps:

mgh = 1/2mV^2 (mass'es cancel)
gh=1/2V^2 (multiply both sides by 2)
2gh=V^2 (square both sides)
V=2gh^1/2

 

[Basic_Physics]

The answer suggest that it was raised to a height of 3/2 h. So the bottom of the block was raised up to the top of the hill then.

 

[SammyS]

Um, when you start a thread it says (without the forward / in the beginning):

No. The site assumes that you leave its titles alone and in bold, then after the title you put in your text -- not in bold.

As follows:

Homework Statement

A block of mass m and height h slides without friction up a hill rise of height (3/2)h as shown. In order to make it to the top of the hill, the block must have a minimum initial speed of:

Homework Equations

Pe+Ke = Pe(f) + Ke(f)

The Attempt at a Solution

I tried figuring out what the velocity of the block would be if it had slid down the hill. I was assuming it would require that amount of velocity to go back up the hill (maybe I assumed wrong). The answer listed is (3gh)^1/2 However my answer was (2gh)^1/2. The block has a height h but I don't understand why the size of the block should matter. Nor do I have any idea of how to incorporate that fact into my equation, please help.

So, the site is asking me to put it in bold, that's why I did it.

I fully understand why you did post in bold.

In any case here is my algebra in steps:

mgh = 1/2mV^2 (mass'es cancel)
gh=1/2V^2 (multiply both sides by 2)
2gh=V^2 (square both sides)
V=2gh^1/2

The hill has a rise of (3/2)h so that should be

mg((3/2)h) = 1/2mV^2 ,

so all will work out fine.

 

[vysero]

Right I noticed what I did wrong when I was going to sleep, I must have been tired. Thanks for the help.