Frictionless horizontal then ramp?

[conniebear14]

Homework Statement

An object of mass m and initial speed v0 sliding on a horizontal frictionless surface encounters a frictionless ramp. To what vertical height will the object rise before coming to rest?

Homework Equations

.5mv^2 = .5mgh

The Attempt at a Solution

I used above equation and got (v^2)/g = h
by canceling out .5m on each side. I have no clue if that's right...

 

[tiny-tim]

hi conniebear14!

An object of mass m and initial speed v0 sliding on a horizontal frictionless surface encounters a frictionless ramp. To what vertical height will the object rise before coming to rest?
…
.5mv^2 = .5mgh

it would help you to get the correct equation if you stated (in words) what principle you are using, or at least what terms you are using

what is the LHS of your equation supposed to be?

what is the RHS of your equation supposed to be?

 

[conniebear14]

I don't know what that means (lhs and rhs). That was the only equation I could think of. The horizontal plus a ramp is what is messing me up. Which equations would you recommend?

 

[tiny-tim]

I don't know what that means (lhs and rhs).

left-hand side, right-hand side! (of an equation)

your LHS is the KE, 1/2 mv²

your RHS should be the PE, which is not 1/2 mgh, it's … ? ​

 

[conniebear14]

okay so PE is mgdeltaY?
Where do I go from there?
I set .5mv^2 = mgdeltaY
and got deltaY = (v^2)/2g

 

[tiny-tim]

yes, that looks fine